Exercise 5. 4[br][br]1. (a) Define conditional trigonometric identities with example.[br]Solution:[br]The trigonometric identities which are true only under certain conditions are known as conditional trigonometric identities. Examples of conditional identities are:[br](i) If [math]A + B + C = \pi^c [/math] then [math] \sin (A + B) = \sin C [/math][br](ii) If [math] A + B + C = \pi^c [/math] then [math] \tan (A + B) = – \tan C [/math][br](iii) If [math] A + B = 90^{\circ} [/math] then [math] \sin A = cos B [/math][br][br]1. (b) What is the true condition for the identity [math] \tan A = \cot B [/math] ?[br]Solution:[br] [math] \tan A = \cot B [/math] is true, only when [math] A+ B = \frac{\pi^c}{2} [/math][br]1. (c) Write any three relations which can be formed from [math]A + B + C = \pi^c [/math][br]Solution:[br]Any three relations that can be formed from [math]A + B + C = \pi^c [/math] are[br][math] \begin{align}[br]\ & (i) \ A+B = \pi^c - C \\[br]\ & (ii) \ \frac{A}{2} + \frac{B}{2} = \frac{\pi^c}{2} - \frac{C}{2} \\[br]\ & (iii) \ 2A+2B = 2\pi^c - 2C \\[br]\end{align} [/math][br][br]2. (a) If [math] A+B+C = \pi^c [/math], prove that: [math] \tan A + \tan B + \tan C = \tan A \tan B \tan C [/math][br]Solution:[br][math] \begin{align} \text{ Here }, \ & \\[br]\ & A+B+C =\pi^c\\[br]or, \ & A+B =\pi^c -C \\ [br]\ & \text{Taking tan on both sides }\\[br]or, \ & \tan (A+B) =\tan(\pi^c -C)\\[br]or, \ & \frac{\tan A + \tan B }{1- \tan A \tan B} = -\tan C\\[br]or, \ & \tan A + \tan B = -\tan C( 1-\tan A \tan B)\\[br]or, \ & \tan A + \tan B = -\tan C + \tan A \tan B \tan C\\[br]\therefore \ & \tan A +\tan B + \tan C =\tan A \tan B \tan C \\[br]\end{align} [/math][br]2. (b) If [math] A+B+C = \pi^c [/math], prove that: [math] \tan \frac{A}{2} \tan \frac{B}{2} +\tan \frac{B}{2} \tan \frac{C}{2} +\tan \frac{C}{2} \tan \frac{A}{2} =1 [/math][br]Solution:[br][math] \begin{align} \text{ Here }, \ & \\[br]\ & A+B+C =\pi^c\\[br]or, \ & \frac{A}{2} + \frac{B}{2} +\frac{C}{2}=\frac{ \pi^c}{2}\\[br]or, \ & \frac{A}{2} + \frac{B}{2} =\frac{ \pi^c}{2}-\frac{C}{2}\\ [br]\ & \text{Taking tan on both sides }\\[br]or, \ & \tan \left(\frac{A}{2} + \frac{B}{2}\right) =\tan \left(\frac{ \pi^c}{2}-\frac{C}{2}\right)\\[br]or, \ & \frac{\tan \frac{A}{2} + \tan \frac{B }{2}}{1- \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2} \\[br]or, \ & \frac{\tan \frac{A}{2} + \tan \frac{B }{2}}{1- \tan \frac{A}{2} \tan \frac{B}{2}} = \frac{1}{\tan \frac{C}{2} }\\[br]or, \ & \tan \frac{C}{2}\left(\tan \frac{A}{2} + \tan \frac{B }{2} \right)= 1- \tan \frac{A}{2} \tan \frac{B}{2} \\[br]or, \ & \tan \frac{C}{2} \tan \frac{A}{2} + \tan \frac{B}{2} \tan \frac{C}{2} = 1- \tan \frac{A}{2} \tan \frac{B}{2} \\[br]\therefore \ & \tan \frac{A}{2} \tan \frac{B}{2} +\tan \frac{B}{2} \tan \frac{C}{2} +\tan \frac{C}{2} \tan \frac{A}{2} =1 \\[br]\end{align} [/math][br][br]2. (c) If [math] A+B+C = \pi^c [/math], prove that: [math] \tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C [/math][br]Solution:[br][math] \begin{align} \text{ Here }, \ & \\[br]\ & 2A+2B+2C =\pi^c\\[br]or, \ & 2A+2B =2\pi^c -2C \\ [br]\ & \text{Taking tan on both sides }\\[br]or, \ & \tan (2A+2B) =\tan(2\pi^c -2C)\\[br]or, \ & \frac{\tan 2A + \tan 2B }{1- \tan 2A \tan 2B} = -\tan 2C\\[br]or, \ & \tan 2A + \tan 2B = -\tan 2C( 1-\tan 2A \tan 2B)\\[br]or, \ & \tan 2A + \tan 2B = -\tan 2C + \tan 2A \tan 2B \tan 2C\\[br]\therefore \ & \tan 2A +\tan 2B + \tan 2C =\tan 2A \tan 2B \tan 2C \\[br]\end{align} [/math][br][br][br]2. (d) If [math] A+B+C = \pi^c [/math], prove that: [math] \cot A \cot B + \cot B \cot C + \cot C \cot A -1 =0 [/math][br]Solution:[br][math] \begin{align} \text{ Here }, \ & \\[br]\ & A+B+C =\pi^c\\[br]or, \ & A+B =\pi^c -C \\ [br]\ & \text{Taking cot on both sides }\\[br]or, \ & \cot (A+B) =\cot (\pi^c -C)\\[br]or, \ & \frac{\cot A \cot B -1}{\cot A + \cot B} = -\cot C\\[br]or, \ & \cot A \cot B -1 = -\cot C( \cot A + \cot B)\\[br]or, \ & \cot A \cot B -1 = -\cot C \cot A - \cot B \cot C\\[br]\therefore \ & \cot A \cot B + \cot B \cot C + \cot C \cot A -1 =0 \\[br]\end{align} [/math][br][br]2. (e) If [math] A+B+C = \pi^c [/math], prove that: [math] \cot 2A \cot 2B + \cot 2B \cot2C + \cot 2C \cot 2A = 1 [/math][br]Solution:[br][math] \begin{align} \text{ Here }, \ & \\[br]\ & A +B+C =\pi^c \\[br]or, \ & 2A+2B+2C =2\pi^c\\[br]or, \ & 2A+2B =2\pi^c -2C \\ [br]\ & \text{Taking cot on both sides }\\[br]or, \ & \cot (2A+2B) =\cot (2\pi^c -2C)\\[br]or, \ & \frac{\cot 2A \cot 2B -1}{\cot 2A + \cot 2B} = -\cot 2C\\[br]or, \ & \cot 2A \cot 2B -1 = -\cot 2C( \cot 2A + \cot 2B)\\[br]or, \ & \cot 2A \cot 2B -1 = -\cot 2C \cot 2A - \cot 2B \cot 2C\\[br]\therefore \ & \cot 2A \cot 2B + \cot 2B \cot 2C + \cot 2C \cot 2A -1 =0 \\[br]\end{align} [/math][br][br]3. (a) If [math] A, B, \text{ and } C [/math] are the vertices of [math] \triangle ABC [/math] then prove that: [math] \sin A + \sin B + \sin C = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin A + \sin B + \sin C\\[br]\ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C\\[br]\ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] \\[br]\ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]3. (b) If [math] A, B, \text{ and } C [/math] are the vertices of [math] \triangle ABC [/math] then prove that: [math] \sin A + \sin B - \sin C = 4\sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin A + \sin B - \sin C\\[br]\ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} - \sin C\\[br]\ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) -2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \frac{C}{2} \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ 2\sin \frac{A}{2} \sin \frac{B}{2}\right] \\[br]\ & = 4\sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}\\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br]3. (c) If [math] A, B, \text{ and } C [/math] are the vertices of [math] \triangle ABC [/math] then prove that: [math] \sin A - \sin B - \sin C = -4\cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin A - \sin B - \sin C\\[br]\ & = 2\cos \frac{A+B}{2} \sin \frac{A-B}{2} - \sin C\\[br]\ & = 2 \cos \left( \frac{A}{2} + \frac{B}{2} \right) \sin \left( \frac{A}{2} + \frac{B}{2} \right) -2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\sin \frac{C}{2} \sin \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \frac{C}{2} \right] \\[br]\ & = 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\[br]\ & = - 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} + \frac{B}{2} \right) -\sin \left( \frac{A}{2} - \frac{B}{2} \right) \right] \\[br]\ & = -2\sin \frac{C}{2} \left[ 2\cos \frac{A}{2} \sin \frac{B}{2}\right] \\[br]\ & = 4\cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]3. (d) If [math] A, B, \text{ and } C [/math] are the vertices of [math] \triangle ABC [/math] then prove that: [math] \cos A + \cos B - \cos C =-1+ 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos A + \cos B - \cos C\\[br]\ & = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2} - \cos C\\[br]\ & = 2 \cos \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} - \frac{B}{2} \right) -\left ( 1-2\sin^2 \frac{C}{2} \right) \\[br]\ & = 2\sin \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) -1 + 2\sin^2 \frac{C}{2} \\ [br]\ & = 2\sin \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin^2 \frac{C}{2} -1 \\ [br]\ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] -1 \\[br]\ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] -1\\[br]\ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} + \frac{B}{2} \right) + \cos \left( \frac{A}{2} - \frac{B}{2} \right) \right] -1 \\[br]\ & = 2\sin \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] -1 \\[br]\ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} -1 \\[br]\ & = -1+ 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]3. (e) If [math] A, B, \text{ and } C [/math] are the vertices of [math] \triangle ABC [/math] then prove that: [math] -\cos A + \cos B + \cos C = 4\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} - 1 [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = - \cos A + \cos B + \cos C\\[br]\ & = \cos B - \cos A + \cos C\\[br]\ & = 2\sin \frac{A+B}{2} \sin \frac{A-B}{2} + \cos C\\[br]\ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \left (2\cos^2 \frac{C}{2} -1 \right) \\[br]\ & = 2\cos \frac{C}{2} \sin \left( \frac{A}{2} - \frac{B}{2} \right) + 2\cos^2 \frac{C}{2} -1 \\[br]\ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \frac{C}{2} \right] -1 \\[br]\ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \left( \frac{A}{2} + \frac{B}{2} \right) \right] - 1 \\[br]\ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} + \frac{B}{2} \right) + \sin \left( \frac{A}{2} - \frac{B}{2} \right) \right] -1 \\[br]\ & = 2\cos \frac{C}{2} \left[ 2\sin \frac{A}{2} \cos \frac{B}{2}\right] -1 \\[br]\ & = 4\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}-1 \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]4. (a) If [math] A+B+C = \pi^c [/math], prove that: [math] \sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & 2A+2B+2C = \pi^c \\[br]\text{or, } \ & 2A+2B+2C = 2\pi^c \\[br]\text{or, } \ & \sin (2A+2B) = \sin(2\pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (2A+2B) = \cos (2\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin 2A + \sin 2B + \sin 2C \\[br]\ & = 2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C\\[br]\ & = 2\sin (A+B) \cos (A-B) + \sin 2C\\[br]\ & = 2\sin C \cos (A-B) + 2\sin C \cos C\\[br]\ & = 2\sin C [ \cos (A-B) + \cos C ]\\[br]\ & = 2\sin C[ \cos (A-B) -\cos (A+B)]\\[br]\ & =2\sin C[ 2\sin A \sin B ]\\[br]\ & = 4\sin A \sin B \sin C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]4. (b) If [math] A+B+C = \pi^c [/math], prove that: [math] \sin 2A - \sin 2B + \sin 2C = 4\cos A \sin B \cos C [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & 2A+2B+2C = \pi^c \\[br]\text{or, } \ & 2A+2B+2C = 2\pi^c \\[br]\text{or, } \ & \sin (2A+2B) = \sin(2\pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (2A+2B) = \cos (2\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin 2A - \sin 2B + \sin 2C \\[br]\ & = 2\cos \frac{2A+2B}{2} \sin\frac{2A-2B}{2} +\sin 2C\\[br]\ & = 2\cos (A+B) \sin (A-B) + \sin 2C\\[br]\ & = -2\cos C \sin (A-B) + 2\sin C \cos C\\[br]\ & = 2\cos C [ -\sin (A-B) + \sin C ]\\[br]\ & = 2\cos C [ \sin C - \sin (A-B) ]\\[br]\ & = 2\cos C[ \sin (A+B) -\sin (A-B)]\\[br]\ & =2\cos C[ 2\cos A \sin B ]\\[br]\ & = 4\cos A \sin B \cos C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]4. (c) If [math] A+B+C = \pi^c [/math], prove that: [math] \sin 2A - \sin 2B - \sin 2C = - 4\sin A \cos B \cos C [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & 2A+2B+2C = \pi^c \\[br]\text{or, } \ & 2A+2B+2C = 2\pi^c \\[br]\text{or, } \ & \sin (2A+2B) = \sin(2\pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (2A+2B) = \cos (2\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin 2A - \sin 2B - \sin 2C \\[br]\ & = 2\cos \frac{2A+2B}{2} \sin\frac{2A-2B}{2} -\sin 2C\\[br]\ & = 2\cos (A+B) \sin (A-B) - \sin 2C\\[br]\ & = -2\cos C \sin (A-B) - 2\sin C \cos C\\[br]\ & = -2\cos C [ \sin (A-B) + \sin C ]\\[br]\ & = -2\cos C [ \sin C + \sin (A-B) ]\\[br]\ & = -2\cos C[ \sin (A+B) + \sin (A-B)]\\[br]\ & =-2\cos C[ 2\sin A \cos B ]\\[br]\ & = -4\sin A \cos B \cos C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br]4. (d) If [math] A+B+C = \pi^c [/math], prove that: [math] \cos 2A - \cos 2B + \cos 2C =1 - 4\sin A \cos B \sin C [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & 2A+2B+2C = \pi^c \\[br]\text{or, } \ & 2A+2B+2C = 2\pi^c \\[br]\text{or, } \ & \sin (2A+2B) = \sin(2\pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (2A+2B) = \cos (2\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos 2A - \cos 2B + \cos 2C \\[br]\ & = -2\sin \frac{2A+2B}{2} \sin\frac{2A-2B}{2} + 1-2\sin^2 C\\[br]\ & = -2\sin (A+B) \sin (A-B) -2\sin^2 C+1 \\[br]\ & = -2\sin C \sin (A-B) - 2\sin^2 C+1 \\[br]\ & = -2\sin C [ \sin (A-B) + \sin C ]+ 1\\[br]\ & = -2\sin C[ \sin (A+B) + \sin (A-B)]+1\\[br]\ & =-2\sin C[ 2\sin A \cos B ] +1 \\[br]\ & = -4\sin A \cos B \sin C +1 \\[br]\ & = 1-4\sin A \cos B \sin C\\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]4. (e) If [math] A+B+C = \pi^c [/math], prove that: [math] \cos 2A + \cos 2B - \cos 2C =1 - 4\sin A \sin B \cos C [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & 2A+2B+2C = \pi^c \\[br]\text{or, } \ & 2A+2B+2C = 2\pi^c \\[br]\text{or, } \ & \sin (2A+2B) = \sin(2\pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (2A+2B) = \cos (2\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos 2A + \cos 2B - \cos 2C \\[br]\ & = 2\cos \frac{2A+2B}{2} \cos \frac{2A-2B}{2} -( 2\cos^2 C -1)\\[br]\ & = 2\cos (A+B) \cos (A-B) - 2\cos^2 C + 1\\[br]\ & = -2\cos C \cos (A-B) - 2\cos^2 C +1\\[br]\ & = -2\cos C [ \cos (A-B) + \cos C ] + 1\\[br]\ & = -2\cos C [ \cos (A-B) - \cos (A+B) ] + 1\\[br]\ & =-2\cos C[ 2\sin A \sin B ] + 1 \\[br]\ & = -4\sin A \sin B \cos C +1 \\[br]\ & = 1-4\sin A \sin B \cos C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]5. (a) If [math] A+B+C = \pi^c [/math], prove that: [math] \sin (B+C-A) + \sin (C+A-B) + \sin (A+B-C) =4\sin A \sin B \sin C [/math][br]Solution:[br][math] \begin{align}[br]\ & A+B+C=\pi^c \\[br]\ & A+B =\pi^c-C ...(i)\\[br]\ & B+C = \pi^c -A ...(ii)\\[br]\ & C+A =\pi^c - B ...(iii) \\[br]\text{Now, }\ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & =\sin (B+C-A) + \sin (C+A-B) + \sin (A+B-C) \\[br]\ & = \sin (\pi^c -A-A) + \sin (\pi^c - B - B) + \sin (\pi^c-C -C) \\[br]\ & = \sin (\pi^c -2A) + \sin (\pi^c - 2B) + \sin (\pi^c-2C) \\[br]\ & = \sin 2A + \sin 2B + \sin 2C \\[br]\ & = 2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C\\[br]\ & = 2\sin (A+B) \cos (A-B) + \sin 2C\\[br]\ & = 2\sin C \cos (A-B) + 2\sin C \cos C\\[br]\ & = 2\sin C [ \cos (A-B) + \cos C ]\\[br]\ & = 2\sin C[ \cos (A-B) -\cos (A+B)]\\[br]\ & =2\sin C[ 2\sin A \sin B ]\\[br]\ & = 4\sin A \sin B \sin C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]5. (b) If [math] A+B+C = \pi^c [/math], prove that: [math] \cos (B+C-A) + \cos (C+A-B) + \cos (A+B-C) =4\cos A \cos B \cos C +1 [/math][br]Solution:[br][math] \begin{align}[br]\ & A+B+C=\pi^c \\[br]\ & A+B =\pi^c-C ...(i)\\[br]\ & B+C = \pi^c -A ...(ii)\\[br]\ & C+A =\pi^c - B ...(iii) \\[br]\text{Now, }\ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & =\cos (B+C-A) + \cos (C+A-B) + \cos (A+B-C) \\[br]\ & = \cos (\pi^c -A-A) + \cos (\pi^c - B - B) + \cos (\pi^c-C -C) \\[br]\ & = \cos (\pi^c -2A) + \cos (\pi^c - 2B) + \cos (\pi^c-2C) \\[br]\ & = - \cos 2A - \cos 2B - \cos 2C \\[br]\ & = - (\cos 2A + \cos 2B) - \cos 2C \\[br]\ & = -2 \cos \frac{2A+2B}{2} \cos \frac{2A-2B}{2} - \cos 2C \\[br]\ & =- 2\cos (A+B) \cos (A-B) - \cos 2C \\[br]\ & = 2\cos C \cos (A-B) - (2\cos^2 C -1 )\\[br]\ & = 2\cos C \cos (A-B) - 2\cos^2 C + 1 \\[br]\ & = 2\cos C [ \cos (A-B) - \cos C ]+ 1\\[br]\ & = 2\cos C[ \cos (A-B) + \cos (A+B)]+ 1 \\[br]\ & =2\cos C[ 2\cos A \cos B ] + 1 \\[br]\ & = 4\cos A \cos B \cos C + 1 \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]5. (c) If [math] A+B+C = \pi^c [/math], prove that: [math] \frac{\cos A}{\sin B \sin C} +\frac{\cos B }{\sin C \sin A } + \frac{\cos C}{\sin B \sin B} =2 [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{\cos A}{\sin B \sin C} +\frac{\cos B }{\sin C \sin A } + \frac{\cos C}{\sin B \sin B}\\[br]\ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C }\\[br]\ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C }\times \frac{2}{2} \\[br]\ & = \frac{ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C}{2\sin A \sin B \sin C }\\[br]\ & = \frac{\sin 2A + \sin 2B + \sin 2C}{ 2\sin A \sin B \sin C}\\[br]\ & = \frac{2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C }{ 2\sin A \sin B \sin C}\\[br]\ & = \frac{2\sin (A+B) \cos (A-B) + \sin 2C}{ 2\sin A \sin B \sin C}\\[br]\ & = \frac{ 2\sin C \cos (A-B) + 2\sin C \cos C}{2\sin A \sin B \sin C}\\[br]\ & = \frac{ 2\sin C [ \cos (A-B) + \cos C ] }{2\sin A \sin B \sin C}\\[br]\ & = \frac{ 2\sin C [ \cos (A-B) - \cos (A+B)] }{2\sin A \sin B \sin C}\\[br]\ & = \frac{ 2\sin C [ 2\sin A \sin B ] }{2\sin A \sin B \sin C}\\[br]\ & = 2 \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]5. (d) If [math] A+B+C = \pi^c [/math], prove that: [math] \frac{\sin A}{\cos B \cos C}+\frac{\sin B}{\cos C \cos A} +\frac{\cos C}{\cos A \cos B } = 2\tan A \tan B \tan C [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{\sin A}{\cos B \cos C}+\frac{\sin B}{\cos C \cos A} +\frac{\cos C}{\cos A \cos B }\\[br]\ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\cos A \cos B \cos C }\\[br]\ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\cos A \cos B \cos C}\times \frac{2}{2} \\[br]\ & = \frac{ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C}{2 \cos A \cos B \cos C}\\[br]\ & = \frac{\sin 2A + \sin 2B + \sin 2C}{ 2\cos A \cos B \cos C}\\[br]\ & = \frac{2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C }{ 2\cos A \cos B \cos C}\\[br]\ & = \frac{2\sin (A+B) \cos (A-B) + \sin 2C}{ 2\cos A \cos B \cos C}\\[br]\ & = \frac{ 2\sin C \cos (A-B) + 2\sin C \cos C}{2\cos A \cos B \cos C}\\[br]\ & = \frac{ 2\sin C [ \cos (A-B) + \cos C ] }{2\cos A \cos B \cos C}\\[br]\ & = \frac{ 2\sin C [ \cos (A-B) - \cos (A+B)] }{2\cos A \cos B \cos C}\\[br]\ & = \frac{ 2\sin C [ 2\sin A \sin B ] }{2\cos A \cos B \cos C}\\[br]\ & = 2\frac{ \sin A }{\cos A} \frac{\sin B}{\cos B} \frac{\sin C}{\cos C}\\[br]\ & = 2\tan A \tan B \tan C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]6. (a) If [math] A+B+C = \pi^c [/math], prove that: [math]\cos^2 A + \cos^2 B +\cos^2 C = 1-2\cos A \cos B \cos C [/math][br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos^2 A + \cos^2 B +\cos^2 C\\[br]\ & = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos^2 C\\[br]\ & = \frac{1+ \cos 2A + 1+ \cos 2B}{2} + \cos^2 C\\[br]\ & = \frac{ 2+ \cos 2A + \cos 2B}{2} + \cos^2 C\\[br]\ & = \frac{ 2+ 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} + \cos^2 C\\[br]\ & = \frac{2+ 2\cos (A+B)\cos (A-B)}{2} + \cos^2 C\\[br]\ & = 1+ \cos (A+B) \cos (A-B) + \cos^2 C\\[br]\ & = 1 - \cos C \cos (A-B) + \cos^2 C \\[br]\ & = 1- \cos C [ \cos (A-B) - \cos C ]\\[br]\ & = 1- \cos C [ \cos (A-B) + \cos (A+B) ] \\[br]\ & = 1 - \cos C [2\cos A \cos B ]\\[br]\ & = 1 - 2\cos A \cos B \cos C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br][br]6. (b) If [math] A+B+C = \pi^c [/math], prove that: [math]\cos^2 A + \cos^2 B -\sin^2 C = -2\cos A \cos B \cos C [/math][br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos^2 A + \cos^2 B - \sin^2 C\\[br]\ & = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} - \sin^2 C\\[br]\ & = \frac{1+ \cos 2A + 1+ \cos 2B}{2} - \sin^2 C\\[br]\ & = \frac{ 2+ \cos 2A + \cos 2B}{2} - \sin^2 C\\[br]\ & = \frac{ 2+ 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} - ( 1 - \cos^2 C )\\[br]\ & = \frac{2+ 2\cos (A+B)\cos (A-B)}{2} - 1 + \cos^2 C )\\[br]\ & = \cos (A+B) \cos (A-B) + \cos^2 C\\[br]\ & = - \cos C \cos (A-B) + \cos^2 C \\[br]\ & = - \cos C [ \cos (A-B) - \cos C ]\\[br]\ & = - \cos C [ \cos (A-B) + \cos (A+B) ] \\[br]\ & = - \cos C [2\cos A \cos B ]\\[br]\ & = - 2\cos A \cos B \cos C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]6. (c) If [math] A+B+C = \pi^c [/math], prove that: [math]\sin^2 A + \sin^2 B + \sin^2 C = 2(1+ \cos A \cos B \cos C ) [/math][br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin^2 A + \sin^2 B + \sin^2 C \\[br]\ & = \frac{1- \cos 2A}{2} + \frac{1- \cos 2B}{2} + \sin^2 C\\[br]\ & = \frac{1- \cos 2A + 1- \cos 2B}{2} + \sin^2 C\\[br]\ & = \frac{ 2 -( \cos 2A + \cos 2B)}{2} + \sin^2 C\\[br]\ & = \frac{ 2 - 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} + ( 1 - \cos^2 C )\\[br]\ & = \frac{2- 2\cos (A+B)\cos (A-B)}{2} + 1 - \cos^2 C )\\[br]\ & = 1- \cos (A+B) \cos (A-B) + 1 - \cos^2 C )\\[br]\ & = 2 + \cos C \cos (A-B) - \cos^2 C \\[br]\ & = 2+ \cos C [ \cos (A-B) - \cos C ]\\[br]\ & = 2+ \cos C [ \cos (A-B) + \cos (A+B) ] \\[br]\ & = 2+ \cos C [2\cos A \cos B ]\\[br]\ & = 2+ 2\cos A \cos B \cos C \\[br]\ & = 2( 1+ \cos A \cos B \cos C ) \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]6. (d) If [math] A+B+C = \pi^c [/math], prove that: [math]\sin^2 A - \sin^2 B - \sin^2 C = - \cos A \sin B \sin C [/math][br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin^2 A - \sin^2 B - \sin^2 C \\[br]\ & = \frac{1- \cos 2A}{2} - \frac{1- \cos 2B}{2} - \sin^2 C\\[br]\ & = \frac{1- \cos 2A - (1- \cos 2B)}{2} - \sin^2 C\\[br]\ & = \frac{1- \cos 2A - 1+\cos 2B}{2} - \sin^2 C\\[br]\ & = \frac{ \cos 2B - \cos 2A}{2} - \sin^2 C\\[br]\ & = \frac{2\sin \frac{2A+2B}{2} \sin \frac{2A-2B}{2} }{2} - \sin^2 C\\[br]\ & = \frac{2\sin (A+B) \sin (A-B) }{2} - \sin^2 C \\[br]\ & = \sin (A+B) \sin (A-B) - \sin^2 C \\[br]\ & = \sin C \sin (A-B) - \sin^2 C\\[br]\ & = \sin C [ \sin (A-B) - \sin C ]\\[br]\ & = \sin C [ \sin (A - B) - \sin (A+B) ]\\[br]\ & = - \sin C [ \sin (A + B) - \sin (A - B) ]\\[br]\ & = - \sin C [ 2\cos A \sin B ]\\[br]\ & = -2\cos A \sin B \sin C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]6. (e) If [math] A+B+C = \pi^c [/math], prove that: [math]\sin^2 A - \sin^2 B + \sin^2 C = \sin A \cos B \sin C [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin^2 A - \sin^2 B + \sin^2 C \\[br]\ & = \frac{1- \cos 2A}{2} - \frac{1- \cos 2B}{2} + \sin^2 C\\[br]\ & = \frac{1- \cos 2A - (1- \cos 2B)}{2} + \sin^2 C\\[br]\ & = \frac{1- \cos 2A - 1+\cos 2B}{2} + \sin^2 C\\[br]\ & = \frac{ \cos 2B - \cos 2A}{2} + \sin^2 C\\[br]\ & = \frac{2\sin \frac{2A+2B}{2} \sin \frac{2A-2B}{2} }{2} + \sin^2 C\\[br]\ & = \frac{2\sin (A+B) \sin (A-B) }{2} + \sin^2 C \\[br]\ & = \sin (A+B) \sin (A-B) + \sin^2 C \\[br]\ & = \sin C \sin (A-B) + \sin^2 C\\[br]\ & = \sin C [ \sin (A-B) + \sin C ]\\[br]\ & = \sin C [ \sin (A - B) + \sin (A+B) ]\\[br]\ & = \sin C [ \sin (A + B) + \sin (A - B) ]\\[br]\ & = \sin C [ 2\sin A \cos B ]\\[br]\ & = 2\sin A \cos B \sin C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]7. (a) If [math] \alpha + \beta + \gamma = 180^{\circ} [/math], prove that:[br][math] \sin^2\frac{\alpha}{2}-\sin^2\frac{\beta}{2}+\sin^2\frac{\gamma}{2} = 1 - 2\cos \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\gamma}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & \alpha + \beta + \gamma = 180^{\circ}\\[br]\text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\[br]\text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\[br]\text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin^2\frac{\alpha}{2}-\sin^2\frac{\beta}{2}+\sin^2\frac{\gamma}{2} \\[br]\ & = \frac{1- \cos \alpha }{2} - \frac{1- \cos \beta }{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{1- \cos \alpha - (1- \cos \beta )}{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{1- \cos \alpha - 1+ \cos \beta )}{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{ \cos \beta - \cos \alpha}{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{2\sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} }{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{2\sin \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) }{2} + \sin^2\frac{\gamma}{2} \\[br]\ & = \sin \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + 1 - \cos^2\frac{\gamma}{2} \\[br]\ & = 1+ \cos \frac{\gamma}{2} \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \cos^2\frac{\gamma}{2} \\[br]\ & = 1+ \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \cos \frac{\gamma}{2} \right] \\[br]\ & = 1+ \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) \right] \\[br]\ & = 1- \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) - \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \right] \\[br]\ & = 1- \cos \frac{\gamma}{2} \left[2 \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \right]\\[br]\ & = 1 - 2\cos \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\gamma}{2}\\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br][br]7. (b) If [math] \alpha + \beta + \gamma = 180^{\circ} [/math], prove that:[br][math] \sin^2\frac{\alpha}{2}+\sin^2\frac{\beta}{2}-\sin^2\frac{\gamma}{2} = 1 - 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & \alpha + \beta + \gamma = 180^{\circ}\\[br]\text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\[br]\text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\[br]\text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin^2\frac{\alpha}{2}+\sin^2\frac{\beta}{2} -\sin^2\frac{\gamma}{2} \\[br]\ & = \frac{1- \cos \alpha }{2} + \frac{1- \cos \beta }{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{1- \cos \alpha + 1 - \cos \beta }{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{ 2 - \cos \beta - \cos \alpha}{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{ 2 - (\cos \beta + \cos \alpha)}{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{2-2\cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} }{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = \frac{2-2\cos \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) }{2} - \sin^2\frac{\gamma}{2} \\[br]\ & = 1 -\cos \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin^2\frac{\gamma}{2} \\[br]\ & = 1 -\sin \frac{\gamma}{2} \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin^2\frac{\gamma}{2} \\[br]\ & = 1- \sin \frac{\gamma}{2} \left[ \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + \sin \frac{\gamma}{2} \right] \\[br]\ & = 1 - \sin \frac{\gamma}{2} \left[ \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + \cos \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) \right] \\[br]\ & = 1- \sin \frac{\gamma}{2} \left[2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\[br]\ & = 1 - 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2}\\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]7. (c) If [math] \alpha + \beta + \gamma = 180^{\circ} [/math], prove that:[br][math] -\cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}+\cos^2\frac{\gamma}{2} = 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & \alpha + \beta + \gamma = 180^{\circ}\\[br]\text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\[br]\text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\[br]\text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = -\cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = -\frac{1+ \cos \alpha }{2} + \frac{1+ \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{- (1+ \cos \alpha) + 1 + \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{ - 1 - \cos \alpha + 1 + \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{ \cos \beta - \cos \alpha}{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{2\sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} }{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} + \cos^2\frac{\gamma}{2} \\[br]\ & = \sin \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos^2\frac{\gamma}{2} \\[br]\ & = \cos \frac{\gamma}{2} \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos^2\frac{\gamma}{2} \\[br]\ & = \cos \frac{\gamma}{2} \left[ \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos \frac{\gamma}{2} \right]\\[br]\ & = \cos \frac{\gamma}{2} \left[ \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\sin \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \right] \\[br]\ & = \cos \frac{\gamma}{2} \left[ 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\[br]\ & = 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]7. (d) If [math] \alpha + \beta + \gamma = 180^{\circ} [/math], prove that:[br][math] \cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}-\cos^2\frac{\gamma}{2} = 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & \alpha + \beta + \gamma = 180^{\circ}\\[br]\text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\[br]\text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\[br]\text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\[br]\text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}-\cos^2\frac{\gamma}{2} \\[br]\ & = \frac{1+ \cos \alpha }{2} + \frac{1+ \cos \beta }{2} - \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{ 1 + \cos \alpha + 1 + \cos \beta }{2} - \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{ 2+ \cos \alpha + \cos \beta}{2} - \cos^2\frac{\gamma}{2} \\[br]\ & = \frac{2+ 2\cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} }{2} - \cos^2\frac{\gamma}{2} \\[br]\ & = 1+ \cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} - \cos^2\frac{\gamma}{2} \\[br]\ & = 1+ \cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) - \left(1-\sin^2\frac{\gamma}{2}\right) \\[br]\ & = 1+ \sin \frac{\gamma}{2} \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) - 1 + \sin^2\frac{\gamma}{2} \\[br]\ & = \sin \frac{\gamma}{2} \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) + \sin^2\frac{\gamma}{2} \\[br]\ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\sin \frac{\gamma}{2} \right]\\[br]\ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \right] \\[br]\ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right)+\cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) \right] \\[br]\ & = \sin \frac{\gamma}{2} \left[ 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\[br]\ & = 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]8. (a) If [math] A+B+C = \pi^c [/math], prove that: [math]\frac{\sin 2A + \sin 2B + \sin 2C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{\sin 2A + \sin 2B + \sin 2C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin \frac{2A + 2B }{2} \cos \frac{2A-2B}{2}+ \sin 2C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin (A + B) \cos (A- B)+ \sin 2C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \cos (A- B)+ 2\sin C \cos C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \left[ \cos (A- B)+ \cos C \right]}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \left[ \cos (A- B) - \cos (A+B) \right]}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \left[ 2\sin A \sin B \right]}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{4 \sin A \sin B \sin C }{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{ \sin A \sin B \sin C }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{\left(2 \sin \frac{A}{2} \cos \frac{A}{2} \right) \left(2\sin \frac{B}{2} \cos \frac{B}{2} \right) \left(2 \sin \frac{C}{2} \cos \frac{C}{2} \right) }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\ [br]\ & = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br][br]8. (b) If [math] A+B+C = \pi^c [/math], prove that: [math]\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A +\sin B + \sin C } = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A +\sin B + \sin C } \\[br]\ & = \frac{2 \sin \frac{2A + 2B }{2} \cos \frac{2A-2B}{2}+ \sin 2C}{2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C } \\[br]\ & = \frac{2 \sin (A + B) \cos (A- B)+ \sin 2C}{ 2\sin \left( \frac{A}{2}+\frac{B}{2} \right)\cos \left( \frac{A}{2}-\frac{B}{2} \right) +\sin C } \\[br]\ & = \frac{2 \sin C \cos (A- B)+ 2\sin C \cos C}{2\cos \frac{C}{2} \cos \left( \frac{A}{2}-\frac{B}{2} \right) + 2\sin \frac{C}{2} \cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \left[ \cos (A- B)+ \cos C \right]}{2\cos \frac{C}{2} \cos \left( \frac{A}{2}-\frac{B}{2} \right) + 2\sin \frac{C}{2} \cos \frac{C}{2} } \\[br]\ & = \frac{2 \sin C \left[ \cos (A- B) - \cos (A+B) \right]}{2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2}-\frac{B}{2} \right) + \sin \frac{C}{2} \right]} \\[br]\ & = \frac{2 \sin C \left[ 2\sin A \sin B \right]}{2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2}-\frac{B}{2} \right) + \cos \left( \frac{A}{2}+ \frac{B}{2} \right) \right] } \\[br]\ & = \frac{2 \sin C \left[ 2\sin A \sin B \right]}{2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{C}{2} \right] } \\[br]\ & = \frac{4 \sin A \sin B \sin C }{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{ \sin A \sin B \sin C }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = \frac{\left(2 \sin \frac{A}{2} \cos \frac{A}{2} \right) \left(2\sin \frac{B}{2} \cos \frac{B}{2} \right) \left( 2\sin \frac{C}{2} \cos \frac{C}{2} \right) }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\[br]\ & = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br][br]8. (c) If [math] A+B+C = \pi^c [/math], prove that: [math]\frac{\sin A + \sin B - \sin C}{\sin A \sin B } = \sec \frac{A}{2} \sec \frac{B}{2} \cos \frac{C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{\sin A + \sin B - \sin C}{\sin A \cos B }\\[br]\ & = \frac{ 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} - \sin C}{ \sin A \sin B }\\[br]\ & = \frac{ 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) -2\sin \frac{C}{2}\cos \frac{C}{2} }{ \sin A \sin B }\\[br]\ & = \frac{ 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin \frac{C}{2}\cos \frac{C}{2} } { \sin A \sin B \sin C }\\[br]\ & = \frac{ 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \frac{C}{2} \right] } { \sin A \sin B }\\[br]\ & = \frac{ 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] } { \sin A \sin B } \\[br]\ & = \frac{ 2\cos \frac{C}{2} \left[ 2\sin \frac{A}{2} \sin \frac{B}{2}\right] } { \sin A \sin B }\\[br]\ & = \frac{ 4\sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} } { 2\sin \frac{A}{2}\cos \frac{A}{2}\times 2\sin \frac{B}{2} \cos \frac{ B}{2} }\\[br]\ & = \frac{\cos \frac{C}{2} }{ \cos \frac{A}{2}\cos \frac{B}{2} } \\[br]\ & = \sec \frac{A}{2} \sec \frac{B}{2} \cos \frac{C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br]8. (d) If [math] A+B+C = \pi^c [/math], prove that: [math]\frac{\sin^ 2A + \sin^ 2B - \sin^ 2C}{\sin A \sin B \sin C } = 2\cot C [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = \pi^c \\[br]\text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\[br]\text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \frac{ \sin^2 A + \sin^2 B - \sin^2 C } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \frac{1- \cos 2A}{2} + \frac{1- \cos 2B}{2} - \sin^2 C } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \frac{1- \cos 2A + 1- \cos 2B }{2} - \sin^2 C } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \frac{ 2-( \cos 2B + \cos 2A) }{2} - \sin^2 C } {\sin A \sin B \sin C } \\[br]\ & = \frac{ \frac{2 - 2\cos \frac{2A+2B}{2} \cos \frac{2A-2B}{2} }{2} - \sin^2 C } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \frac{2- 2\cos (A+B) \cos (A-B) }{2} - (1- \cos^2 C) } {\sin A \sin B \sin C } \\[br]\ & = \frac{ 1- \cos (A+B) \cos (A-B) - 1 + \cos^2 C } {\sin A \sin B \sin C } \\[br]\ & = \frac{ \cos C \cos (A-B) + \cos^2 C } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \cos C [ \cos (A-B) + \cos C ] } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \cos C [ \cos (A - B) - \cos (A+B) ] } {\sin A \sin B \sin C }\\[br]\ & = \frac{ \cos C [ 2\sin A \sin B ] } {\sin A \sin B \sin C }\\[br]\ & = \frac{ 2\sin A \sin B \cos C } {\sin A \sin B \sin C }\\[br]\ & = 2\times \frac{\cos C}{\sin C}\\[br]\ & = 2\cot C \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]9. (a) If [math] X+Y+Z=180^{\circ} [/math], prove that: [math] \sin X \cos Y \cos Z + \sin Y \cos Z \cos X + \sin Z cos X \cos Y = \sin X \sin Y \sin Z [/math][br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & A+B+C = 180^{\circ} \\[br]\text{or, } \ & \sin (X + Y ) = \sin( 180^{\circ} - Z )=\sin Z \\[br]\text{or, } \ & \cos ( X + Y ) = \cos (180^{\circ}- Z )=- \cos Z \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin X \cos Y \cos Z + \sin Y \cos Z \cos X + \sin Z cos X \cos Y \\[br]\ & = \cos Z ( \sin X \cos Y + \cos X \sin Y ) + \sin Z \cos X \cos Y \\[br]\ & = \cos Z \sin (X+Y) + \sin Z \cos X \cos Y \\[br]\ & = \cos Z \sin Z + \sin Z \cos X \cos Y \\[br]\ & = \sin Z ( \cos Z + \cos X \cos Y ) \\[br]\ & = \sin Z ( - \cos (X+Y) + \cos X \cos Y ) \\[br]\ & = \sin Z [ - ( \cos X \cos Y - \sin X \sin Y ) + \cos X \cos Y ] \\[br]\ & = \sin Z [ - \cos X \cos Y + \sin X \sin Y + \cos X \cos Y ] \\[br]\ & = \sin Z [ \sin X \sin Y ] \\[br]\ & = \sin X \sin Y \sin Z \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]9. (b) If [math] X+Y+Z=180^{\circ} [/math], prove that: [math] \cos X \sin Y \sin Z + \cos Y \sin Z \sin X + \cos Z \sin X \sin Y - \cos X \cos Y \cos Z =1 [/math] [br]Solution:[br][math] \begin{align} [br]\text{Given, } \ & \\[br]\ & X+Y+Z = 180^{\circ} \\[br]\text{or, } \ & \sin ( X + Y) = \sin( 180^{\circ} - Z )=\sin Z \\[br]\text{or, } \ & \cos (X+Y) = \cos (180^{\circ}- Z )=- \cos Z \\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos X \sin Y \sin Z + \cos Y \sin Z \sin X + \cos Z \sin X \sin Y - \cos X \cos Y \cos Z \\[br]\ & = \sin Z ( \cos X \sin Y + \cos Y \sin X ) + \cos Z ( \sin X \sin Y - \cos X \cos Y ) \\[br]\ & = \sin Z ( \sin X \cos Y + \cos X \sin Y ) - \cos Z ( \cos X \cos Y - \sin X \sin Y ) \\[br]\ & = \sin Z \sin (X+Y) - \cos Z \cos (X+Y) \\[br]\ & = \sin Z \sin Z - \cos Z ( -\cos Z ) \\[br]\ & = \sin^2 Z + \cos^2 Z \\[br]\ & = 1 \\[br]\ & = \text{RHS}\\ [br]\end{align} [/math][br][br]10. (a) If [math] A+B+C = \pi^c [/math], prove that: [math]\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} = 1 + 4\sin \left(\frac{A+B}{4} \right) \sin \left(\frac{B+C}{4} \right) \sin \left(\frac{C+A}{4} \right)[/math] [br]Solution:[br]Solution:[br][math] \begin{align}[br]\text{Given}\\ \ & A+B+C = \pi^c \\[br]\ & A + B = \pi^c - C \cdots (i) \\[br]\ & B + C = \pi^c - A \cdots (ii) \\[br]\ & C + A = \pi^ c - B \cdots (iii) \\[br]\text{LHS} \ & = \sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \\[br]\ & = 2\sin \frac{\frac{A}{2} + \frac{B}{2}}{2} \cos \frac{\frac{A}{2} - \frac{B}{2}}{2} +\cos \left( \frac{\pi^c}{2} - \frac{C}{2} \right) \\[br]\ & = 2\sin \frac{A+B}{4} \cos \frac{A-B}{4} +\cos \frac{\pi^c-C}{2} \\[br]\ & = 2\sin \frac{\pi^c - C}{4} \cos \frac{A-B}{4} + 1-2\sin^2 \frac{\pi^c-C}{4}\\[br]\ & =1+ 2\sin \frac{\pi^c - C}{4} \cos \frac{A-B}{4} -2\sin^2 \frac{\pi^c-C}{4}\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \sin \frac{\pi^c+C}{4} \right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \left( \frac{\pi^c}{2} - \frac{\pi^c-C}{4} \right) \right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \left( \frac{2\pi^c-(\pi^c - C) }{4} \right) \right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \frac{\pi^c + C }{4} \right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2 \sin \frac{ \frac{A-B}{4} + \frac{\pi^c + C }{4} }{2} \sin \frac{ \frac{\pi^c + C}{4} - \frac{A-B}{4} }{2} \right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ A-B+\pi^c + C }{4\times 2 } \sin \frac{\pi^c+C-A+B }{4\times 2}\right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ (A+C)+ \pi^c - B }{ 8 } \sin \frac{\pi^c -A+(B+C) }{ 8 }\right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - B+ \pi^c - B }{ 8 } \sin \frac{\pi^c -A+\pi^c -A }{ 8 }\right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2\pi^c - 2B }{ 8 } \sin \frac{2\pi^c -2A}{ 8 }\right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2(\pi^c - B ) }{ 8 } \sin \frac{2( \pi^c -A )}{ 8 }\right]\\[br]\ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -A }{ 4 }\right]\\[br]\ & = 1+ 4 \sin \frac{\pi^c -C}{4} \sin \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -A }{ 4 } \\[br]\ & = 1+ 4 \sin \frac{A+B}{4} \sin \frac{B+C }{ 4 } \sin \frac{ C+A}{ 4 } \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]10. (b) If [math] A+B+C = \pi^c [/math], prove that:[math] \cos \frac{A}{2} - \cos \frac{B}{2} + \cos \frac{C}{2} = 4\cos \frac{\pi^c +A}{4}\cos \frac{\pi^c -B}{4}\cos \frac{\pi^c +C}{4} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given}\\[br]\ & A+B+C = \pi^c \\[br]\ & A + B = \pi^c - C \cdots (i) \\[br]\ & B + C = \pi^c - A \cdots (ii) \\[br]\ & C + A = \pi^ c - B \cdots (iii) \\[br]\text{LHS} \ & = \cos \frac{A}{2} - \cos \frac{B}{2} + \cos \frac{C}{2} \\[br]\ & = 2\sin \frac{\frac{A}{2} + \frac{B}{2}}{2} \sin \frac{\frac{B}{2} - \frac{A}{2}}{2} +\sin \left( \frac{\pi^c}{2} - \frac{C}{2} \right) \\[br]\ & = 2\sin \frac{A+B}{4} \sin \frac{B-A}{4} +\sin \frac{\pi^c-C}{2} \\[br]\ & = 2\sin \frac{A+B}{4} \sin \frac{B-A}{4} +\sin 2\left(\frac{\pi^c-C}{4} \right) \\[br]\ & = 2\sin \frac{\pi^c - C}{4} \sin \frac{B-A}{4} + 2\sin \frac{\pi^c-C}{4}\cos \frac{\pi^c-C}{4} \\ [br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \cos \frac{\pi^c - C}{4} \right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{B-A}{4} + \cos \left( \frac{\pi^c}{2} - \frac{\pi^c-C}{4} \right) \right]\\[br]\ & =2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \sin \left( \frac{2\pi^c-(\pi^c - C) }{4} \right) \right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \sin \frac{\pi^c + C }{4} \right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ 2 \sin \frac{ \frac{B-A}{4} + \frac{\pi^c + C }{4} }{2} \cos \frac{ \frac{\pi^c + C}{4} - \frac{B-A}{4} }{2} \right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ B-A+\pi^c + C }{4\times 2 } \cos \frac{\pi^c+C - B+A }{4\times 2}\right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c -A + ( B+C) }{ 8 } \cos \frac{\pi^c -B+(A+C) }{ 8 }\right]\\[br]\ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - A+ \pi^c - A }{ 8 } \cos \frac{\pi^c -B+\pi^c -B }{ 8 }\right]\\[br]\ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2\pi^c - 2A }{ 8 } \cos \frac{2\pi^c -2B}{ 8 }\right]\\[br]\ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2(\pi^c - A ) }{ 8 } \cos \frac{2( \pi^c -B )}{ 8 }\right]\\[br]\ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - A }{ 4 } \cos \frac{ \pi^c -B }{ 4 }\right]\\[br]\ & = 4 \sin \frac{\pi^c -A}{4} \cos \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -C }{ 4 } \\[br]\ & = 4 \cos\left( \frac{\pi^c}{2} - \frac{\pi^c -A}{4} \right) \cos \frac{ \pi^c - B }{ 4 } \cos \left( \frac{ \pi^c }{2} - \frac{ \pi^c -C }{ 4 }\right) \\[br]\ & = 4 \cos \frac{ 2\pi^c - \pi^c + A}{4} \cos \frac{\pi^c - B}{4} \cos \frac{ 2\pi^c - \pi^c + C } { 4} \\[br]\ & = 4 \cos \frac{ \pi^c + A}{4} \cos \frac{\pi^c - B}{4} \cos \frac{ \pi^c + C } { 4} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br][br]10. (c) If [math] A+B+C = \pi^c [/math], prove that: [math]\sin A + \sin B +\sin C = 4\sin \frac{B+C}{2}\sin \frac{C+A}{2}\sin \frac{A+B}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = \pi^c \\ [br]\text{or, } \ & \frac{A+B+C }{2} =\frac{\pi^c}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = \frac{\pi^c}{2}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(\frac{\pi^c}{2}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(\frac{\pi^c}{2}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \sin A + \sin B + \sin C\\[br]\ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C\\[br]\ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\[br]\ & = 2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] \\[br]\ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\[br]\ & = 4\sin \left( \frac{ \pi^c}{2} - \frac{A}{2} \right)\sin \left( \frac{ \pi^c}{2} - \frac{B}{2} \right)\sin \left( \frac{ \pi^c}{2} - \frac{C}{2} \right) \\[br]\ & = 4\sin \frac{\pi^c - A}{2} \sin \frac{\pi^c - B}{2} \sin \frac{\pi^c - C}{2} \\ [br]\ & = 4\sin \frac{B+C}{2}\sin \frac{C+A}{2}\sin \frac{A+B}{2}[ \ \ \because A + B + C = \pi^c \ \ ] \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br]10. (d) If [math] A+B+C = \pi^c [/math], prove that: [math] \cos A + \cos B + \cos C = 1+ 4\cos \frac{\pi^c -A }{2}\cos \frac{\pi^c - B}{2}\cos \frac{\pi^c - C}{2} [/math][br]Solution:[br][math] \begin{align}[br]\text{Given } \ & \\[br]\ & A+B+C = 180^{\circ}\\[br]\text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\[br]\text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\[br]\text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\[br]\text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\[br]\text{Now} \ & \\[br]\text{LHS} \ & = \cos A + \cos B + \cos C\\[br]\ & = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2} + \cos C\\[br]\ & = 2 \cos \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 1-2\sin^2 \frac{C}{2} \\[br]\ & = 2\sin \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin^2 \frac{C}{2} +1 \\ [br]\ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \frac{C}{2} \right] + 1 \\[br]\ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] + 1\\[br]\ & = 2\sin \frac{C}{2} \left[ 2\sin \frac{A}{2} \sin \frac{B}{2}\right] + 1 \\[br]\ & = 4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} + 1 \\[br]\ & = 1+ 4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\[br]\ & = 1+ 4\cos \left(\frac{\pi^c}{2} - \frac{A}{2} \right)\cos \left(\frac{\pi^c}{2} - \frac{B}{2} \right)\cos \left(\frac{\pi^c}{2} - \frac{C}{2} \right)\\[br]\ & = 1+ 4\cos \frac{\pi^c - A}{2} \cos \frac{\pi^c - B}{2} \cos \frac{\pi^c - C}{2} \\[br]\ & = \text{RHS}\\[br]\end{align} [/math][br][br] The End