Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & x = {\text{cosec}}\theta - \sin \theta {\text{ }} \cr & y = \sec \theta - \cos \theta \cr & {\text{Put }}\theta = {45^ \circ } \cr & x = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & y = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & {\text{by options (B) }}{x^2}{y^2}\left( {{x^2} + {y^2} + 3} \right) \cr & = {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \left[ {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 3} \right] \cr & = \frac{1}{2} \times \frac{1}{2}\left( {\frac{1}{2} + \frac{1}{2} + 3} \right) \cr & = \frac{1}{4}\left( {1 + 3} \right) \cr & = 1{\text{ }}\left( {{\text{satisfy}}} \right) \cr} $$

Solution:

 $$\eqalign{ & \frac{x}{8} = \sin \theta + \cos \theta ........\left( {\text{i}} \right) \cr & \frac{y}{9} = \sin \theta - \cos \theta ........\left( {{\text{ii}}} \right) \cr & {\text{Square and add equation}}\left( {\text{i}} \right){\text{and}}\left( {{\text{ii}}} \right) \cr & \frac{{{x^2}}}{{{8^2}}} + \frac{{{y^2}}}{{{9^2}}} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta + {\sin ^2}\theta + {\cos ^2}\theta - 2\sin \theta \cos \theta \cr & \frac{{{x^2}}}{{{8^2}}} + \frac{{{y^2}}}{{{9^2}}} = 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{If A + B}} = {90^ \circ } \cr & {\text{Then,}}{\sin ^2}{\text{A + }}{\sin ^2}{\text{B}} = 1 \cr & \Rightarrow 7{\frac{1}{2}^ \circ } + 82{\frac{1}{2}^ \circ } \cr & \Rightarrow {90^ \circ } \cr & \Rightarrow 1\left[ {{\text{sin }}{{90}^ \circ } = 1} \right] \cr} $$

Solution:

 $$\eqalign{ & {\cos ^2}\theta = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & {\text{Max value of }}{\cos ^2}\theta = 1 \cr & \Rightarrow 1 = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & \Rightarrow 4xy = {\left( {x + y} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow 0 = {x^2} + {y^2} - 2xy \cr & \Rightarrow 0 = {\left( {x - y} \right)^2} \cr & \Rightarrow 0 = x - y \cr & \Rightarrow x = y \cr} $$

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:

 $$\eqalign{ & \frac{{\cos {{50}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{8}}{0^ \circ }}}{{\sec {{10}^ \circ }}} - 2\cos {50^ \circ } \cdot {\text{cosec}}\,{40^ \circ }{\text{ angle of sum}} = {90^ \circ } \cr & {\text{then,}} \cr & = \frac{{\sin {{40}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{5}}{0^ \circ }}}{{{\text{cosec}}\,{\text{8}}{0^ \circ }}} - 2\sin {40^ \circ } \cdot {\text{cosec}}\,{10^ \circ } \cr & = 1 + 3 - 2 \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{2cos}}\theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{When,}} \cr & ax \mp by = m \cr & {\text{then, }}bx \mp ay = \sqrt {{a^2} + {b^2} - {m^2}} \cr & 2\cos \theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \theta + 2\sin \theta = \sqrt {4 + 1 - \frac{1}{2}} \cr & \Rightarrow \cos \theta + 2\sin \theta = \frac{3}{{\sqrt 2 }} \cr} $$

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:

 $$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$

Solution:

 $$\eqalign{ & \sec A - \cos A \cr & = \frac{1}{{\cos A}} - \cos A \cr & = \frac{{1 - {{\cos }^2}A}}{{\cos A}} \cr & = \frac{{{{\sin }^2}A}}{{\cos A}} \cr & = \tan A.\sin A \cr} $$