Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \tan {10^ \circ }.\tan {15^ \circ }.\tan {75^ \circ }.\tan {80^ \circ } \cr & \Rightarrow \left( {\tan {{10}^ \circ }.\tan {{80}^ \circ }} \right).\left( {\tan {{15}^ \circ }.\tan {{75}^ \circ }} \right) \cr & \Rightarrow 1 \times 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \Rightarrow \sin {\text{ }}7x = \cos 11x \cr & \Rightarrow 7x + 11x = {90^ \circ } \cr & \Rightarrow 18x = {90^ \circ } \cr & \Rightarrow x = {5^ \circ } \cr & \Rightarrow \tan 9x + \cot 9x \cr & \Rightarrow \tan {45^ \circ } + \cot {45^ \circ } \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

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

$$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

= (tan²α + 1 )sin²β
= (tan²45° + 1 )sin²45°
$$ = (1 + 1) {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$$
$$ = 2 \times \frac{1}{2} = 1$$

$$\eqalign{ & \Rightarrow {\text{sin 3A}} = {\text{cos}}\left( {{\text{A}} - {{26}^ \circ }} \right) \cr & \Rightarrow {\text{3A + A}} - {\text{2}}{{\text{6}}^ \circ }{\text{ = }}{90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 4{\text{A = 11}}{6^ \circ } \cr & \Rightarrow {\text{A}} = {29^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = \cos {60^ \circ }\left( {{0^ \circ } < \theta < {{90}^ \circ }} \right) \cr & \Rightarrow \theta + {18^ \circ } + {60^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = {12^ \circ } \cr & \Rightarrow {\text{cos 5}}\theta {\text{ }} \cr & \Rightarrow {\text{cos }}{60^ \circ } \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$

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

$$\eqalign{ & {\left( {\frac{{{\text{sin 4}}{{\text{7}}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\frac{{\cos {{43}^ \circ }}}{{{\text{sin }}{{47}^ \circ }}}} \right)^2} - 4{\text{co}}{{\text{s}}^2}{45^ \circ } \cr & = {\left( {\frac{{{\text{cos 4}}{{\text{3}}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\frac{{sin{{47}^ \circ }}}{{{\text{sin }}{{47}^ \circ }}}} \right)^2} - 4 \times \frac{1}{2} \cr & = 1 + 1 - 2 \cr & = 0 \cr & {\bf{Note:}} \cr & \left( {\sin \left( {{{90}^ \circ } - \theta } \right)} \right) = \cos \theta \cr} $$