Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & \alpha + \beta = {90^ \circ } \cr & {\text{and }}\alpha :\beta = 2:1 \cr & 2x + x = {90^ \circ } \cr & x = {30^ \circ } \cr & {\text{ }}\alpha = {60^ \circ } \cr & \beta = {30^ \circ } \cr & \frac{{{\text{cos }}\alpha }}{{\cos \beta }} \cr & = \frac{{\cos {{60}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \cr & = \frac{1}{{\sqrt 3 }} \cr & = 1:\sqrt 3 \cr} $$

Solution:

 $$\eqalign{ & 3{\sec ^2}\theta + \tan \theta - 7 = 0 \cr & \Rightarrow 3\left( {1 + {{\tan }^2}\theta } \right) + \tan \theta - 7 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + \tan \theta - 4 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + 4\tan \theta - 3\tan \theta - 4 = 0 \cr & \Rightarrow \tan \theta \left( {3\tan \theta + 4} \right) - 1\left( {3\tan \theta + 4} \right) = 0 \cr & \Rightarrow \left( {3\tan \theta + 4} \right)\left( {\tan \theta - 1} \right) = 0 \cr & \Rightarrow \tan \theta = 1\,\,\,\,\,\,\,\,\,\therefore \theta = {45^ \circ } \cr & \therefore \,\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }} \cr & = \frac{{2\left( {\frac{1}{{\sqrt 2 }}} \right) + 3\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 + \sqrt 2 }} \cr & = \frac{{\frac{5}{{\sqrt 2 }}}}{{2\sqrt 2 }} \cr & = \frac{5}{4} \cr} $$

Solution:

Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution:

 $$\eqalign{ & \tan 7\theta .\tan 2\theta = 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \left( {7\theta + 2\theta } \right) = {90^ \circ } \cr & 9\theta = {90^ \circ } \cr & \theta = {10^ \circ } \cr & \Rightarrow \tan 3\theta \cr & \Rightarrow \tan {30^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 3 }} \cr} $$

Solution:

 $$\eqalign{ & \sin 10 - \frac{4}{3}{\sin ^3}10 \cr & = \frac{{3\sin 10 - 4{{\sin }^3}10}}{3} \cr & = \frac{1}{3}\sin 3 \times {10^ \circ }\,\,\,\,\,\left[ {\because \sin A = 3\sin A - 4{{\sin }^3}A} \right] \cr & = \frac{1}{3} \times \frac{1}{2} \cr & = \frac{1}{6} \cr} $$

Solution:

 $$\eqalign{ & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}x - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{1 + 2\left( {{{\sec }^2}x - 1} \right) - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2{{\sec }^2}x - 1 - 2\sec x\tan x}}{{\sec x - \tan x}} \cr & \Rightarrow \frac{{2\sec x\left( {\sec x - \tan x} \right)}}{{\sec x - \tan x}} - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \cr & \Rightarrow 2\sec x - \frac{1}{{\sec x - \tan x}} \times \frac{{\sec x + \tan x}}{{\sec x + \tan x}} \cr & \Rightarrow 2\sec x - \frac{{\sec x + \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}} \cr & \Rightarrow 2\sec x - \sec x - \tan x \cr & \Rightarrow \sec x - \tan x \cr & \cr & {\bf{Alternative:}} \cr & \frac{{1 + 2{{\cot }^2}\left( {{{90}^ \circ } - x} \right) - 2{\text{cosec}}\left( {{{90}^ \circ } - x} \right)\cot \left( {{{90}^ \circ } - x} \right)}}{{{\text{cosec}}\left( {{{90}^ \circ } - x} \right) - \cot \left( {{{90}^ \circ } - x} \right)}} \cr & {\text{By putting }}x = {45^ \circ }{\text{ in equation}} \cr & \Rightarrow \frac{{1 + 2{{\tan }^2}{{45}^ \circ } - 2\sec {{45}^ \circ }\tan {{45}^ \circ }}}{{\sec {{45}^ \circ } - \tan {{45}^ \circ }}} \cr & \Rightarrow \frac{{1 + 2 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{{3 - 2\sqrt 2 }}{{\sqrt 2 - 1}} \times \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \cr & \Rightarrow 3\sqrt 2 - 2\sqrt 2 - 1 \cr & \Rightarrow \sqrt 2 \cr & {\text{By satisfying in options}} \cr & \Rightarrow \sec x - \tan x \Rightarrow \sqrt 2 - 1 \cr} $$

Solution:

 $$\eqalign{ & \cos {20^ \circ } = m{\text{ }} \cr & \cos {70^ \circ } = n \cr & {\text{So,}} \cr & \Leftrightarrow {m^2} + {n^2} = {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & \left[ {{\text{If co}}{{\text{s}}^2}{\text{A + co}}{{\text{s}}^2}{\text{B}} = {\text{1}}} \right] \cr & ({\text{If, A}} + {\text{B}} = {90^ \circ }) \cr & \Leftrightarrow 1 \cr} $$

Solution:

 (sec x sec y + tan x tan y)2 - (sec x tan y + tan x sec y)2 = sec2x. sec2y + tan2x. tan2y + 2sec x. sec y. tan x. tan y - sec2x. tan2y - tan2x. sec2y + 2sec x. tan y. tan x sec y = sec2x [sec2y - tan2y] - tan2x [sec2y - tan2y] = (sec2x - tan2x) (sec2y - tan2y) = 1 × 1 = 1

Solution:

 $$\eqalign{ & {\sin ^2}\theta - 3\sin \theta + 2 = 0 \cr & \Rightarrow {\sin ^2}\theta - 2{\text{sin }}\theta - \sin \theta + 2 = 0 \cr & \Rightarrow \sin \theta \left( {\sin \theta - 2} \right) - 1\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \left( {\sin \theta - 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \left[ {\because \sin \theta \ne 2} \right]{\text{Put value of }} \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow {\text{sin }}\theta = \sin {90^ \circ } \cr & \Rightarrow \theta = {90^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }}\theta = {90^ \circ } \cr & \left[ {{\text{Take help from the options}}} \right] \cr & \Rightarrow {\sin ^2}\theta - 3\sin {\text{ }}\theta + 2 = 0 \cr & \Rightarrow {\sin ^2}{90^ \circ } - 3{\text{sin }}{90^ \circ } + 2 = 0 \cr & \Rightarrow 1 - 3 \times 1 + 2 = 0 \cr & \left[ {\sin {{90}^ \circ } = 1} \right] \cr & \Rightarrow 0 = 0\left[ {{\text{matched}}} \right] \cr & {\text{So, this is answer}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & \cos \theta = \frac{{12}}{{13}} = \frac{B}{H} \cr & P = \sqrt {{{13}^2} - {{12}^2}} = 5\,{\text{cm}} \cr & \frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}} \cr & = \frac{{\frac{P}{H}\left( {1 - \frac{P}{B}} \right)}}{{\frac{P}{B}\left( {1 + \frac{H}{P}} \right)}} \cr & = \frac{{\frac{5}{{13}}\left( {1 - \frac{5}{{12}}} \right)}}{{\frac{5}{{12}}\left( {1 + \frac{{13}}{5}} \right)}} \cr & = \frac{{\frac{5}{{13}} \times \frac{7}{{12}}}}{{\frac{5}{{12}} \times \frac{{18}}{5}}} \cr & = \frac{{\frac{{35}}{{156}}}}{{\frac{3}{2}}} \cr & = \frac{{35 \times 2}}{{156 \times 3}} \cr & = \frac{{35}}{{234}} \cr} $$