Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & x = \frac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }} \cr & {\text{Let's put the value of }}\theta = {45^ \circ } \cr & {\text{So, }}x = \frac{{2 \times \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}} \cr & \left\{ {{\text{Because }}\sin {{45}^ \circ } = \frac{1}{{\sqrt 2 }}{\text{ and}}\cos {{45}^ \circ } = \frac{1}{{\sqrt 2 }}} \right\} \cr & x = \frac{{\sqrt 2 }}{{\frac{{\left( {\sqrt 2 + 2} \right)}}{{\sqrt 2 }}}} \cr & x = \frac{{\sqrt 2 }}{{\left( {1 + \sqrt 2 } \right)}} \cr & {\text{Now, check in the required value}} \cr & \frac{{1 - \cos \theta + \sin \theta }}{{1 + \sin \theta }} \cr & = \frac{{1 - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }}}} \cr & = \frac{{\sqrt 2 }}{{\left( {\sqrt 2 + 1} \right)}} \cr & {\text{This values is same as }}x. \cr & {\text{Hence answer is }}x. \cr} $$

tan4θ = cot(40° - 2θ)
tan4θ = cot(90° - 40° + 2θ)
tan4θ = tan(50° + 2θ)
4θ = 50° + 2θ
2θ = 50°
θ = 25°

$$\eqalign{ & \frac{{32{{\cos }^6}x - 48{{\cos }^4}x + 18{{\cos }^2}x - 1}}{{4\sin x\, \times \cos x \times \,\sin \left( {60 - x} \right) \times \cos \left( {60 - x} \right) \times \sin \left( {60 + x} \right) \times \cos \left( {60 + x} \right)}} \cr & {\text{Adding }} + 4\,{\text{and }} - 4, \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 18{\cos ^2}x + 3 \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 24{\cos ^2}x - 6{\cos ^2}x + 3 \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 24{\cos ^2}x - 3\cos 2x \cr & \Rightarrow 4{\left( {2{{\cos }^2}x - 1} \right)^3} - 3\cos 2x \cr & \left[ {\therefore \,\cos 2x = 2{{\cos }^2}x - 1} \right] \cr & \Rightarrow 4{\cos ^3}2x - 3\cos 2x \cr & \Rightarrow \cos 6x \cr} $$
\[\left[ \begin{align} & \because \,\cos 3x\to 4{{\cos }^{3}}x-3\cos x \\ & \text{and} \\ & \because \,\cos 6x\to 4{{\cos }^{3}}2x-3\cos 2x \\ \end{align} \right]\]
$$\eqalign{ & {\text{Now, }}4\sin x\,.\cos x.\,\sin \left( {60 - x} \right).\cos \left( {60 - x} \right).\sin \left( {60 + x} \right).\cos \left( {60 + x} \right) \cr & \left[ {\therefore \,\sin x.\sin \left( {60 - x} \right).\sin \left( {60 + x} \right) = \frac{1}{4}\sin 3x} \right] \cr & \left[ {\therefore \,\cos x.\cos \left( {60 - x} \right).\cos \left( {60 + x} \right) = \frac{1}{4}\cos 3x} \right] \cr & = \frac{{\cos 6x}}{{\frac{1}{4}\sin 3x.\cos 3x}} \cr & = \frac{2}{2} \times \frac{{4\cos 6x}}{{\sin 3x.\cos 3x}} \cr & = \frac{{8\cos 6x}}{{2\sin 3x.\cos 3x}} \cr & = \frac{{8\cos 6x}}{{\sin 6x}} \cr & = 8\cot 6x \cr} $$

$$\eqalign{ & \left( {\frac{{{{\tan }^3}\theta }}{{{{\sec }^2}\theta }} + \frac{{{{\cot }^3}\theta }}{{{\text{cose}}{{\text{c}}^2}\,\theta }} + 2\sin \theta \cos \theta } \right) \div \left( {1 + {\text{cose}}{{\text{c}}^2}\theta + {{\tan }^2}\theta } \right) \cr & = \left( {\frac{{{{\sin }^3}\theta }}{{\cos \theta }} + \frac{{{{\cos }^3}\theta }}{{{\text{sin}}\,\theta }} + 2\sin \theta \cos \theta } \right) \div \left( {{\text{cose}}{{\text{c}}^2}\theta + {{\sec }^2}\theta } \right) \cr & = \left( {\frac{{{{\sin }^4}\theta + {{\cos }^4}\theta + 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}} \right) \div \left( {\frac{1}{{{{\sin }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right) \cr & = \left( {\frac{{{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^2}}}{{\sin \theta \cos \theta }}} \right) \div \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right) \cr & = \left( {\frac{1}{{\sin \theta \cos \theta }}} \right) \times \left( {\frac{{{{\sin }^2}\theta {{\cos }^2}\theta }}{1}} \right) \cr & = \sin \theta \cos \theta \cr} $$

$$\eqalign{ & \frac{{{{\left( {1 - \sin \theta + \cos \theta } \right)}^2}\left( {1 - \cos \theta } \right){{\sec }^3}\theta {\text{cose}}{{\text{c}}^2}\theta }}{{\left( {\sec \theta - \tan \theta } \right)\left( {\tan \theta + \cot \theta } \right)}} \cr & {\text{put }}\theta = {30^ \circ } \cr & = \frac{{{{\left( {1 - \sin {{30}^ \circ } + \cos {{30}^ \circ }} \right)}^2}\left( {1 - \cos {{30}^ \circ }} \right){{\sec }^3}{{30}^ \circ }{\text{cose}}{{\text{c}}^2}{{30}^ \circ }}}{{\left( {\sec {{30}^ \circ } - \tan {{30}^ \circ }} \right)\left( {\tan {{30}^ \circ } + \cot {{30}^ \circ }} \right)}} \cr & = \frac{{{{\left( {1 - \frac{1}{2} + \frac{{\sqrt 3 }}{2}} \right)}^2}\left( {1 - \frac{{\sqrt 3 }}{2}} \right){{\left( {\frac{2}{{\sqrt 3 }}} \right)}^3}{{\left( 2 \right)}^2}}}{{\left( {\frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }}} \right)\left( {\frac{1}{{\sqrt 3 }} + \sqrt 3 } \right)}} \cr & = \frac{{{{\left( {\frac{{1 + \sqrt 3 }}{2}} \right)}^2}\left( {\frac{{2 - \sqrt 3 }}{2}} \right){{\left( {\frac{2}{{\sqrt 3 }}} \right)}^3}{{\left( 2 \right)}^2}}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)\left( {\frac{{1 + 3}}{{\sqrt 3 }}} \right)}} \cr & = \frac{{\frac{{4 + 2\sqrt 3 }}{4} \times \frac{{2 - \sqrt 3 }}{2} \times \frac{8}{{3\sqrt 3 }} \times 4}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)\left( {\frac{4}{{\sqrt 3 }}} \right)}} \cr & = \frac{{\left[ {{2^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \times \frac{8}{{3\sqrt 3 }}}}{{\frac{4}{3}}} \cr & = \frac{{\frac{8}{{3\sqrt 3 }}}}{{\frac{4}{3}}} \cr & = \frac{2}{{\sqrt 3 }} \cr & {\text{Option A: }}2\tan {30^ \circ } = \frac{2}{{\sqrt 3 }} \cr & {\text{Option B: }}\cot {30^ \circ } = \sqrt 3 \cr & {\text{Option C: }}\sin {30^ \circ } = \frac{1}{2} \cr & {\text{Option D: }}2\cos {30^ \circ } = 2 \times \frac{{\sqrt 3 }}{2} = \sqrt 3 \cr & {\text{Only option A is answer}}{\text{.}} \cr} $$

$$\eqalign{ & \frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{{\text{k}} + 1}}{{{\text{k}} - 1}} \cr & \frac{{\tan \theta }}{{\sin \theta }} = {\text{k}} \cr & {\text{k}} = \sec \theta \cr} $$

$$\eqalign{ & \sin \theta = \frac{{{p^2} - 1}}{{{p^2} + 1}} \cr & {\text{Let }}p = 2 \cr} $$

$$\eqalign{ & {\text{Now from option A}} \cr & \frac{{2 \times 2}}{{1 + 4}} = \frac{4}{5} \cr} $$

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

$$\eqalign{ & m\left[ {\sin \theta + 2{{\cos }^2}\phi + 3\sin \theta + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right] \cr & \theta = {30^ \circ },\,\,\phi = {45^ \circ } \cr & = m\left[ {\left( {\sin \theta + 3\sin \theta + 5\sin \theta + \,.....\,17\sin \theta } \right) + \left( {2{{\cos }^2}\phi + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right)} \right] \cr & \Rightarrow {\text{Sum of odd number}} = {\left( 9 \right)^2}, \cr & {\text{Sum of even number}} = 9 \times 10 \cr & = m\left[ {81\sin \theta + 90{{\cos }^2}\phi } \right] \cr & = m\left[ {81 \times \sin 30 + 90 \times {{\cos }^2}45} \right] \cr & = m\left[ {81 \times \frac{1}{2} + 90 \times \frac{1}{2}} \right] \cr & = m \times \frac{{171}}{2} \cr & {\text{Now, we will check through the option,}} \cr & {\text{Putting }}m = 152 \cr & = 152 \times \frac{{171}}{2} \cr & = 4 \times 19 \times 19 \times 9 \cr & = {\left( {2 \times 19 \times 3} \right)^2} \cr & = {\text{ perfect square}}{\text{. Ans}}{\text{.}} \cr} $$

1 + sin²θ - 3sinθ.cosθ = 0
⇒ 1 - 2sinθ.cosθ = sinθ.cosθ - sin²θ
⇒ sin²θ + cos²θ - 2sinθ.cosθ = sinθ(cosθ - sinθ)
⇒ (cosθ - sinθ)² = sinθ(cosθ - sinθ)
⇒ cosθ - sinθ = sinθ
⇒ cosθ = 2sinθ
⇒ cotθ = 2