Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

tan²θ + cot²θ - sec²θ.cosec²θ
Put θ = 45°
= tan²45° + cot²45° - sec²45°.cosec²45°
= 1 + 1 - (√2)².(√2)²
= 2 - 4
= -2

$$\eqalign{ & 1 + 2{\tan ^2}\theta + 2\sin \theta {\sec ^2}\theta = \frac{a}{b} \cr & 1 + {\tan ^2}\theta + {\tan ^2}\theta + 2\sin \theta \frac{1}{{{{\cos }^2}\theta }} = \frac{a}{b} \cr & {\sec ^2}\theta + {\tan ^2}\theta + 2\tan \theta \sec \theta = \frac{a}{b} \cr & {\left( {\sec \theta + \tan \theta } \right)^2} = \frac{a}{b} \cr & {\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & {\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & \frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 + {{\sin }^2}\theta }} = \frac{a}{b} \cr & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{a}{b} \cr & {\text{Use componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = {\text{cosec}}\,\theta = \frac{{a + b}}{{a - b}} \cr} $$

$$\eqalign{ & \sin \theta = 4\cos \theta \cr & \tan \theta = 4 \cr & \sin \theta .\cos \theta = \frac{4}{{\sqrt {17} }} \times \frac{1}{{\sqrt {17} }} = \frac{4}{{17}} \cr} $$

$$\eqalign{ & \frac{{2\left( {1 - {{\sin }^2}\theta } \right){\text{cose}}{{\text{c}}^2}\theta }}{{{{\cot }^2}\theta \left( {1 + {{\tan }^2}\theta } \right)}} - 1 \cr & = \frac{{2{{\cos }^2}\theta {\text{cose}}{{\text{c}}^2}\theta }}{{{{\cot }^2}\theta {{\sec }^2}\theta }} - 1 \cr & = \frac{{2{{\cos }^2}\theta {{\sin }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta {{\sec }^2}\theta }} - 1 \cr & = 2{\cos ^2}\theta - 1 \cr & = \cos 2\theta \cr} $$

$$\eqalign{ & \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - A} \right) + \cot \left( {{{90}^ \circ } - A} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2A} \right)} \right]}} \cr & {\text{By putting }}A = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right) + \cot \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2 \times {{45}^ \circ }} \right)} \right]}} \cr & \Rightarrow \frac{{{{\left[ {\tan {{45}^ \circ } + \cot {{45}^ \circ }} \right]}^2}}}{{2{{\sec }^2}\left( {{{90}^ \circ } - {{90}^ \circ }} \right)}} \cr & \Rightarrow \frac{{{{\left[ {1 + 1} \right]}^2}}}{{2{{\sec }^2}{0^ \circ }}} \cr & \Rightarrow \frac{4}{{2 \times 1}} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1 \cr & = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{\left( {\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)^2} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

$$\eqalign{ & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = \frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}} \cr & \sin \theta = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}} \cr} $$

$$\eqalign{ & AB = \sqrt {{{\left( {{p^2} + {q^2}} \right)}^2} - {{\left( {{p^2} - {q^2}} \right)}^2}} \cr & AB = \sqrt {4{p^2}{q^2}} \cr & AB = 2pq \cr & \sec \theta = \frac{{{p^2} + {q^2}}}{{2pq}} \cr & \sec \theta = \frac{1}{2}\left[ {\frac{p}{q} + \frac{q}{p}} \right] \cr} $$

$$\eqalign{ & \frac{{\left[ {\sin \left( {90 - A} \right) + \cos \left( {180 - 2A} \right)} \right]}}{{\left[ {\cos \left( {90 - 2A} \right) + \sin \left( {180 - A} \right)} \right]}} \cr & \Rightarrow \frac{{\cos A + \left( { - \cos 2A} \right)}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{\cos A - \cos 2A}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{ - 2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{ - \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{{2A - A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{A}{2}} \right)}}{{\cos \left( {\frac{A}{2}} \right)}} \cr & \Rightarrow \tan \left( {\frac{A}{2}} \right) \cr} $$

$$\eqalign{ & \cot {53^ \circ } = \frac{x}{y} = \frac{B}{H} \cr & {P^2} = {H^2} - {B^2} \cr & P = \sqrt {{y^2} - {x^2}} \cr & \sec {53^ \circ } + \cot {37^ \circ } \cr & = \sec {53^ \circ } + \tan {53^ \circ } \cr & = \frac{H}{B} + \frac{P}{B} \cr & = \frac{{H + P}}{B} \cr & = \frac{{y + \sqrt {{y^2} - {x^2}} }}{x} \cr} $$