Trigonometric Ratio and Identities MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & \left( {\sec \,\alpha + \tan \,\alpha } \right)\left( {\sec \,\beta + \tan \,\beta } \right)\left( {\sec \,\gamma + \tan \,\gamma } \right) = \,\tan \,\alpha \,\tan \,\beta \,\tan \,\gamma \cr & \Rightarrow \,\left( {{{\sec }^2}\,\alpha - {{\tan }^2}\alpha } \right)\left( {{{\sec }^2}\beta - {{\tan }^2}\beta } \right)\left( {{{\sec }^2}\gamma - {{\tan }^2}\gamma } \right) \cr & = \tan \alpha \,\tan\beta \,\tan \gamma \left( {\sec \alpha - \tan \alpha } \right)\left( {\sec \beta - \tan\beta } \right)\left( {\sec \gamma - \tan \gamma } \right) \cr & \Rightarrow \,\left( {\sec \alpha - \tan \alpha } \right)\left( {\sec \beta - \tan\beta } \right)\left( {\sec \gamma - \tan \gamma } \right) \cr & = \cot \alpha \cot \beta \,\cot \gamma \cr} $$

$$\eqalign{ & \cot{5^ \circ }\,\cot {10^ \circ }\,.....\cot {85^ \circ } \cr & = \cot {5^ \circ }\,\cot {10^ \circ }.....\cot \left( {{{90}^ \circ } - {{10}^ \circ }} \right)\cot \left( {{{90}^ \circ } - {5^ \circ }} \right) \cr & = \cot {5^ \circ }\,\cot {10^ \circ }.....\tan {10^ \circ }\tan {5^ \circ } \cr & = \left( {\tan {5^ \circ }\,\cot {5^ \circ }} \right)\left( {\tan {{10}^ \circ }\,\cot {{10}^ \circ }} \right)..... \cr & = \left( 1 \right)\left( 1 \right)\left( 1 \right)..... = 1 \cr} $$

Value $$ = \frac{{\sqrt 3 \,{\text{cos}}\,{{20}^ \circ } - \sin {{20}^ \circ }}}{{\sin {{20}^ \circ } \cdot \cos {{20}^ \circ }}} = \frac{{4\left( {\frac{{\sqrt 3 }}{2}\cos {{20}^ \circ } - \frac{1}{2}\sin {{20}^ \circ }} \right)}}{{\sin {{40}^ \circ }}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4\sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)}}{{\sin {{40}^ \circ }}} = 4.$$

Given that $$\sin \theta = \frac{1}{2}\,{\text{and}}\,\cos \phi = \frac{1}{3}\,{\text{and }}\theta \,{\text{and }}\phi $$        both are acute angles
$$\eqalign{ & \therefore \,\,\theta = \frac{\pi }{6}\,{\text{and 0 < }}\frac{1}{3} < \frac{1}{2} \cr & {\text{or }}\cos \frac{\pi }{2} < \cos \phi < \cos \frac{\pi }{3}\,{\text{or }}\frac{\pi }{3} < \phi < \frac{\pi }{2} \cr & \therefore \,\,\frac{\pi }{3} + \frac{\pi }{6} < \theta + \phi < \frac{\pi }{2} + \frac{\pi }{6}{\text{ or }}\frac{\pi }{2} < \theta + \phi < \frac{{2\pi }}{3} \cr & \Rightarrow \,\,\theta + \phi \in \left( {\frac{\pi }{2},\frac{{2\pi }}{3}} \right) \cr} $$

$$\eqalign{ & \pi < \alpha - \beta < 3\pi \cr & \Rightarrow \,\,\frac{\pi }{2} < \frac{{\alpha - \beta }}{2} < \frac{{3\pi }}{2} \cr & \Rightarrow \,\,\cos \frac{{\alpha - \beta }}{2} < 0 \cr & \sin \alpha + \sin \beta = - \frac{{21}}{{65}} \cr & \Rightarrow \,\,2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2} = - \frac{{21}}{{65}}\,\,\,\,\,.....\left( 1 \right) \cr & \cos \alpha + \cos \beta = - \frac{{27}}{{65}} \cr & \Rightarrow \,\,2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2} = - \frac{{27}}{{65}}\,\,\,\,.....\left( 2 \right) \cr} $$
Square and add (1) and (2)
$$\eqalign{ & 4{\cos ^2}\frac{{\alpha - \beta }}{2} \cr & = \frac{{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}}}{{{{\left( {65} \right)}^2}}} \cr & = \frac{{1170}}{{65 \times 65}} \cr & \therefore \,\,{\cos ^2}\frac{{\alpha - \beta }}{2} \cr & = \frac{9}{{130}} \cr & \Rightarrow \,\,\cos \frac{{\alpha - \beta }}{2} \cr & = - \frac{3}{{\sqrt {130} }} \cr} $$