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

Here, $${\cos ^4}\theta \cdot {\sec ^2}\alpha + {\sin ^4}\theta \cdot {\text{cose}}{{\text{c}}^2}\alpha = 1$$
$$\eqalign{ & {\text{or, }}{\cos ^4}\theta \cdot {\sin ^2}\alpha + {\sin ^4}\theta \cdot {\cos ^2}\alpha = {\sin ^2}\alpha \cdot {\cos ^2}\alpha \cr & {\text{or, }}\left( {1 - {{\sin }^2}\theta } \right){\cos ^2}\theta \cdot {\sin ^2}\alpha + {\sin ^4}\theta \cdot \left( {1 - {{\sin }^2}\alpha } \right) = {\sin ^2}\alpha \left( {1 - {{\sin }^2}\alpha } \right) \cr & {\text{or, }}{\cos ^2}\theta \,{\sin ^2}\alpha + {\sin ^4}\theta - {\sin ^2}\theta \,{\sin ^2}\alpha \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = {\sin ^2}\alpha - {\sin ^4}\alpha \cr & {\text{or, }}{\sin ^4}\theta + {\sin ^4}\alpha - 2{\sin ^2}\theta \cdot {\sin ^2}\alpha = 0 \cr & \Rightarrow \,\,{\sin ^2}\theta = {\sin ^2}\alpha \,\,{\text{and so, }}{\cos ^2}\theta = {\cos ^2}\alpha . \cr & \therefore \,\,{\cos^8}\theta \cdot {\sec ^6}\alpha + {\sin ^8}\theta \cdot {\text{cose}}{{\text{c}}^6}\alpha = {\cos ^2}\theta + {\sin ^2}\theta = 1. \cr} $$

$$\eqalign{ & \because \sin {18^ \circ } = \frac{{\sqrt 5 - 1}}{4} \cr & {x^2} = {4^2} - {\left( {\sqrt 5 - 1} \right)^2} \cr & \Rightarrow x = \sqrt {10 + 2\sqrt 5 } \cr & \Rightarrow \cos {18^ \circ } = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4} \cr & \Rightarrow 2{\cos ^2}9 - 1 = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4} \cr & {\cos ^2}9 = \frac{{\sqrt {10 + 2\sqrt 5 } + 4}}{8} \cr & \Rightarrow \,{\sin ^2}81 = \frac{{4 + \sqrt {10 + 2\sqrt 5 } }}{8} \cr} $$
After squaring all the options available, we come to a conclusion that option $$\left( A \right)$$  is correct.

$$\eqalign{ & \sin \theta = 3\sin \left( {\theta + 2\alpha } \right) \cr & \Rightarrow \sin \left( {\theta + \alpha - \alpha } \right) = 3\sin \left( {\theta + \alpha + \alpha } \right) \cr & \Rightarrow \sin \left( {\theta + \alpha } \right)\cos \alpha - \cos \left( {\theta + \alpha } \right)\sin \alpha = 3\sin \left( {\theta + \alpha } \right)\cos \alpha + 3\cos \left( {\theta + \alpha } \right)\sin \alpha \cr & \Rightarrow - 2\sin \left( {\theta + \alpha } \right)\cos \alpha = 4\cos \left( {\theta + \alpha } \right)\sin \alpha \cr & \Rightarrow \frac{{ - \sin \left( {\theta + \alpha } \right)}}{{\cos \left( {\theta + \alpha } \right)}} = \frac{{2\sin \alpha }}{{\cos \alpha }} \cr & \Rightarrow \tan\left( {\theta + \alpha } \right) + 2\tan \alpha = 0 \cr} $$

Angle traced by the hour hand in 12 hours $$ = {360^ \circ }$$
Angle traced by it in $$4\,hr\,30\min \left( {4\,h + \frac{{30}}{{60}}hr} \right) = \frac{9}{2}hr$$
$$ = \frac{9}{2} \times \frac{{360}}{{12}} = {135^ \circ }$$
Angle traced by minute hand is $$60\min = {360^ \circ }$$
Angle traced by it in $$30\min = \frac{{30}}{{60}} \times 360 = {180^ \circ }$$
Required Angle $$ = {180^ \circ } - {135^ \circ } = {45^ \circ }$$
$$ \Rightarrow 45 \times \frac{\pi }{{180}} = \frac{\pi }{4}{\text{radian}}$$

Given that $$\alpha + \beta = \frac{\pi }{2}\,$$
$$\eqalign{ & \Rightarrow \,\alpha = \frac{\pi }{2} - \beta \cr & \Rightarrow \,\,\tan \alpha = \tan \left( {\frac{\pi }{2} - \beta } \right) = \cot \beta = \frac{1}{{\tan \beta }} \cr & \Rightarrow \,\,\tan \alpha \tan \beta = 1 \cr & \Rightarrow \,\,1 + \tan \alpha \tan \beta = 2. \cr & \,\therefore \,\,\tan \left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }} \cr & \Rightarrow \,\,\tan \gamma = \frac{{\tan \alpha - \tan \beta }}{2} \cr & \Rightarrow \,\,2\tan \gamma = \tan \alpha - \tan \beta \cr & \Rightarrow \,\,\tan \alpha = 2\tan \gamma + \tan \beta \cr} $$

$$\eqalign{ & {\text{Let}}\,A = {30^ \circ } \cr & \Rightarrow \sin A + 2\sin 2A + \sin 3A \cr & = \sin {30^ \circ } + 2\sin {60^ \circ } + \sin {90^ \circ } \cr & = \frac{1}{2} + \frac{{2\sqrt 3 }}{2} + 1 = \frac{{2\sqrt 3 + 3}}{2}\left( {\because 2\,{{\cos }^2}A = 1 + \cos 2A} \right) \cr & {\text{Now, }}4\sin 2A{\cos ^2}\left( {\frac{A}{2}} \right) = 2\sin 2A\left[ {1 + \cos A} \right] \cr & = 2\sin {60^ \circ }\left[ {1 + \cos {{30}^ \circ }} \right] = \frac{{2\sqrt 3 + 3}}{2} \cr & {\text{Also, }}\sin 2A = 2\sin A\cos A\,\,\& \,\,{\sin ^2}A + {\cos ^2}A = 1 \cr & 2\sin 2A{\left[ {\sin \frac{A}{2} + \cos \frac{A}{2}} \right]^2} \cr & = 2\sin 2A\left[ {{{\sin }^2}\frac{A}{2} + {{\cos }^2}\frac{A}{2} + 2\sin \frac{A}{2}\cos \frac{A}{2}} \right] \cr & = 2\sin 2A\left[ {1 + \sin A} \right] = 2\sin {60^ \circ }\left[ {1 + \sin {{30}^ \circ }} \right] = \frac{{3\sqrt 3 }}{2} \cr & \& \,\,8\sin A\cos A{\cos ^2}\left( {\frac{A}{2}} \right) \cr & = 4\sin A\cos A\left[ {1 + \cos A} \right] \cr & = 4\sin {30^ \circ }\cos {30^ \circ }\left[ {1 + \cos {{30}^ \circ }} \right] \cr & = \frac{{2\sqrt 3 + 3}}{2} \cr} $$

Value $$ = 2\tan {50^ \circ } - \left( {\tan {{70}^ \circ } - \tan {{20}^ \circ }} \right)$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\tan {50^ \circ } - \frac{{\sin {{50}^ \circ }}}{{\cos {{70}^ \circ } \cdot \cos {{20}^ \circ }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\tan {50^ \circ } - \frac{{2\sin {{50}^ \circ }}}{{2\sin {{20}^ \circ } \cdot \cos {{20}^ \circ }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\tan {50^ \circ } - 2 \cdot \frac{{\sin {{50}^ \circ }}}{{\sin {{40}^ \circ }}} = 2\tan {50^ \circ } - 2 \cdot \frac{{\sin {{50}^ \circ }}}{{\cos {{50}^ \circ }}} = 0. \cr} $$

$$\eqalign{ & x = 1 + {\cos ^2}\phi + {\cos ^4}\phi + .....\,{\text{to }}\infty = \frac{1}{{1 - {{\cos }^2}\phi }} = \frac{1}{{{{\sin }^2}\phi }}. \cr & {\text{Similarly, }}y = \frac{1}{{1 - {{\sin }^2}\phi }} = \frac{1}{{{{\cos }^2}\phi }},z = \frac{1}{{1 - {{\cos }^2}\phi \cdot {{\sin }^2}\phi }} \cr & \therefore \,\,xyz = \frac{1}{{{{\sin }^2}\phi \cdot {{\cos }^2}\phi \left( {1 - {{\sin }^2}\phi \,{{\cos }^2}\phi } \right)}} \cr & xyz = \frac{{\left( {1 - {{\sin }^2}\phi \cdot {{\cos }^2}\phi } \right) + {{\sin }^2}\phi \cdot {{\cos }^2}\phi }}{{{{\sin }^2}\phi \cdot {{\cos }^2}\phi \left( {1 - {{\sin }^2}\phi \,{{\cos }^2}\phi } \right)}} \cr & xyz = \frac{1}{{{{\sin }^2}\phi \,{{\cos }^2}\phi }} + \frac{1}{{1 - {{\sin }^2}\phi \,{{\cos }^2}\phi }} = xy + z. \cr} $$

$$\tan \theta = - \frac{4}{3}$$
$$ \Rightarrow \,\,\theta \in $$  II quad or IV quad.
$$\eqalign{ & \therefore \,\,\,0 < \sin \theta < 1\,\,{\text{or }} - 1 < \sin \theta < 0 \cr & \therefore \,\,\sin \theta \,\,{\text{may be }}\frac{4}{5}\,{\text{or }} - \frac{4}{5} \cr} $$

The direction cosines of the line are $$\cos \theta ,\cos \beta ,\cos \theta $$
$$\eqalign{ & \therefore \,\,{\cos ^2}\theta + {\cos ^2}\beta + {\cos ^2}\theta = 1 \cr & \Rightarrow \,\,2{\cos ^2}\theta = {\sin ^2}\beta = 3{\sin ^2}\theta \,\,\left( {{\text{given}}} \right) \cr & \Rightarrow \,\,2{\cos ^2}\theta = 3 - 3{\cos ^2}\theta \cr & \therefore \,\,{\cos ^2}\theta = \frac{3}{5} \cr} $$