Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & \pi \sin \theta = 1\,.....(i) \cr & \pi \cos \theta = 1\,.....(ii) \cr & {\text{Divide eq}}{\text{. (i) from (ii)}} \cr & \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr & \tan \theta = 1 \cr & \tan \theta = \tan {45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr & = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr & = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr & = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & \cot \theta = 4 \cr & \therefore \frac{{5\sin \theta + 3\cos \theta }}{{5\sin \theta - 3\cos \theta }} \cr & = \frac{{5 + 3\cot \theta }}{{5 - 3\cot \theta }} \cr & = \frac{{5 + 3 \times 4}}{{5 - 3 \times 4}} \cr & = - \frac{{17}}{7} \cr} $$

Solution:

 $$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$   $$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$       $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$   $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$ $$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {60^ \circ }{\text{option A}} \cr & \Rightarrow 2{\sin ^2}{60^ \circ } = 3\cos {60^ \circ } \cr & \Rightarrow 2{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\left( {\frac{1}{2}} \right) \cr & \Rightarrow \frac{3}{2} = \frac{3}{2}{\text{ }}\left( {{\text{LHS}} = {\text{RHS}}} \right) \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{\tan \theta }} + \tan \theta \cr & {\text{put }}\theta = {45^ \circ } \cr & \frac{1}{1} + 1 = 2 \cr & {\text{In option}} \cr & \Rightarrow \left( {\text{A}} \right)\frac{{{\text{cosec}}\,\theta }}{{\sec \theta }} = \frac{{{\text{cosec}}\,{{45}^ \circ }}}{{\sec {{45}^ \circ }}} = \frac{{\sqrt 2 }}{{\sqrt 2 }} = 1 \cr & \Rightarrow \left( {\text{B}} \right)\sec \theta \times {\text{cosec}}\,\theta \cr & = \sec {45^ \circ } \times {\text{cosec}}\,{45^ \circ } \cr & = \sqrt 2 \times \sqrt 2 = 2 \cr & \Rightarrow \left( {\text{C}} \right)\,1 \cr & \Rightarrow \left( {\text{D}} \right){\tan ^2}\theta = {\tan ^2}{45^ \circ } = 1 \cr} $$

Solution:

 $${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$   $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$      $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$    $${\text{cot1}}{5^ \circ }$$ $$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$

Solution:

 $$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$

Solution:

 $$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

Solution:

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