Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\frac{{{{\left( {\tan {{20}^ \circ }} \right)}^2}}}{{{{\left( {{\text{cosec 7}}{0^ \circ }} \right)}^2}}}$$   $$ + $$ $$\frac{{{{\left( {\cot {{20}^ \circ }} \right)}^2}}}{{{{\left( {{\text{sec 7}}{0^ \circ }} \right)}^2}}}$$   $$ + $$ $$2\tan {15^ \circ }$$ . $$\tan {45^ \circ }$$ . $$\tan {75^ \circ }$$ $$\eqalign{ & \Rightarrow \frac{{{{\left( {\tan {{20}^ \circ }} \right)}^2}}}{{{\text{se}}{{\text{c}}^2}{{20}^ \circ }}} + \frac{{{{\left( {\cot {{20}^ \circ }} \right)}^2}}}{{{\text{cose}}{{\text{c}}^2}{{20}^ \circ }}} + 2\tan {15^ \circ }.\tan {75^ \circ } \cr & \left[ {{\text{tan 1}}{5^ \circ }{\text{.tan 7}}{5^ \circ } = {\text{1}}{\text{. If, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{20}^ \circ } + {\text{co}}{{\text{s}}^2}{{20}^ \circ }} \right) + 2 \cr & \Rightarrow 1 + 2 \cr & \Rightarrow 3 \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr & \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow 3\left( 1 \right) \cr & \Rightarrow 3 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\tan }^6}\theta - {{\sec }^6}\theta + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{{{\left( {{{\tan }^2}\theta } \right)}^3} - {{\left( {{{\sec }^2}\theta } \right)}^3} + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right] \cr & = \frac{{\left( {{{\tan }^2}\theta - {{\sec }^2}\theta } \right)\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta + 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta - 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta - 2{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1}}{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)}^2}}} \cr & = - {\left( {\cos \theta \sin \theta } \right)^2} \cr & = - {\cos ^2}\theta {\sin ^2}\theta \cr} $$

Solution:

 $$\eqalign{ & {\text{sin}}\left( {2x - {{20}^ \circ }} \right) = {\text{cos}}\left( {2y + {{20}^ \circ }} \right) \cr & \Rightarrow \left( {2x - {{20}^ \circ }} \right) + \left( {2y + {{20}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B, then A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2\left( {x + y} \right) = {90^ \circ } \cr & \Rightarrow x + y = {45^ \circ } \cr & \therefore \tan \left( {x + y} \right) \cr & = \tan {45^ \circ } \cr & = 1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\sec ^2}{28^ \circ } - {\cot ^2}{62^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = {\sec ^2}{28^ \circ } - {\tan ^2}{28^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = 1 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( 2 \right)^2} \cr & = 1 + \frac{3}{4} + 4 \cr & = \frac{{23}}{4} \cr} $$

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:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$

Solution:

 $$\eqalign{ & \sqrt 3 \tan \theta = 3\sin \theta \cr & \Rightarrow \frac{{\tan \theta }}{{\sin \theta }} = \frac{3}{{\sqrt 3 }} \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} \times \frac{1}{{\sin \theta }} = \frac{3}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\cos \theta }} = \sqrt 3 \cr & \Rightarrow \cos \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow {\cos ^2}\theta = \frac{1}{3} \cr & \therefore \,{\sin ^2}\theta - {\cos ^2}\theta \cr & = 1 - {\cos ^2}\theta - {\cos ^2}\theta \cr & = 1 - 2{\cos ^2}\theta \cr & = 1 - 2 \times \frac{1}{3} \cr & = \frac{1}{3} \cr} $$

Solution:

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