Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{1 + }}\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}} - {\text{se}}{{\text{c}}^2}{27^ \circ }{\text{ + }}\frac{1}{{{{\sin }^2}{{63}^ \circ }}} - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}{63^ \circ } - {\text{se}}{{\text{c}}^2}{27^ \circ } + {\text{cose}}{{\text{c}}^2}{63^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\sec ^2}{27^ \circ } + {\text{se}}{{\text{c}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 2 }}{\text{sin}}\frac{\pi }{6}{\text{.cos}}\frac{\pi }{4} - \cot \frac{\pi }{3}{\text{.sec}}\frac{\pi }{6}{\text{ + }}\frac{{5\tan \frac{\pi }{4}}}{{12\sin \frac{\pi }{2}}} \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} \times \frac{2}{{\sqrt 3 }} + \frac{{5 \times 1}}{{12 \times 1}} \cr & \Rightarrow \frac{1}{4} - \frac{2}{3} + \frac{5}{{12}} \cr & \Rightarrow \frac{{3 - 8 + 5}}{{12}} \cr & \Rightarrow 0 \cr & {\text{ }} \cr} $$

$${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$
$$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$
$$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$

$$\eqalign{ & {\text{ta}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{t}}^2}{\text{A}} - {\sec ^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}{45^ \circ } + {\text{co}}{{\text{t}}^2}{45^ \circ } - {\sec ^2}{45^ \circ }{\text{.cose}}{{\text{c}}^2}{45^ \circ } \cr & \Rightarrow 1 + 1 - {\left( {\sqrt 2 } \right)^2}{\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow - 2 \cr} $$

$$\eqalign{ & {\text{sin}}\left( {4\alpha - \beta } \right) = 1 = \sin {90^ \circ } \cr & 4\alpha - \beta = {90^ \circ }{\text{ }} \cr & {\text{cos}}\left( {2\alpha + \beta } \right) = \frac{1}{2} = \cos {60^ \circ } \cr & 2\alpha + \beta = {60^ \circ } \cr & {\text{Adding }}6\alpha = {150^ \circ } \cr & \alpha = {25^ \circ } \cr & \beta = {10^ \circ } \cr & \Rightarrow {\text{sin}}\left( {\alpha + 2\beta } \right) \cr & \Rightarrow {\text{sin}}\left( {{{25}^ \circ } + 2 \times {{10}^ \circ }} \right) \cr & \Rightarrow {\text{sin}}{45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & {\text{4co}}{{\text{s}}^2}\theta - 1 = 0 \cr & \Rightarrow {\text{4co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & \Rightarrow \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \theta = {60^ \circ } \cr & \Rightarrow \tan \left( {\theta - {{15}^ \circ }} \right) \cr & \Rightarrow \tan \left( {{{60}^ \circ } - {{15}^ \circ }} \right) \cr & \Rightarrow \tan {45^ \circ } \cr & \Rightarrow 1 \cr} $$

$$\frac{{\sin {{25}^ \circ }.\cos {{65}^ \circ } + \cos {{25}^ \circ }.\sin {{65}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - \cos e{c^2}{{20}^ \circ }}}$$
$$ = \frac{{\sin {{25}^ \circ }.\cos \left( {{{90}^ \circ } - {{25}^ \circ }} \right) + \,\cos {{25}^ \circ }.\,\sin \left( {{{90}^ \circ } - {{25}^ \circ }} \right)}}{{{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}$$
$$\eqalign{ & = \frac{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - se{c^2}{{70}^ \circ }}} \cr & = \frac{1}{{ - 1}} \cr & = - 1 \cr} $$

$$\eqalign{ & \theta + \phi = \frac{\pi }{2} \cr & \theta + \phi = {90^ \circ }\,......(i) \cr & \sin \theta = \frac{1}{2} \cr & \sin \theta = {\text{sin 3}}{0^ \circ } = \frac{1}{2}\,......(ii) \cr & {\text{Put }}\theta = {30^ \circ }{\text{ in equation (i)}} \cr & {30^ \circ } + \phi = {90^ \circ } \cr & \phi = {60^ \circ } \cr & \sin \phi = \sin {60^ \circ } = \frac{{\sqrt 3 }}{2} \cr} $$

$$\eqalign{ & {\text{The value of }} \cr & {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) + 2\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) + 12{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta \cr & {\text{Using }}\theta = {0^ \circ } \cr & \because \sin {0^ \circ } = {0^ \circ } \cr & \cos {0^ \circ } = 1 \cr & \Rightarrow 3\left( {0 + {1^4}} \right) + 2\left( {0 + {1^6}} \right) + 12 \times 0 \times 1 \cr & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}}{\text{ + }}\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ }}}{{{{\sin }^2}{{45}^ \circ }}} - \frac{{{\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{45}^ \circ }}} - \frac{{{{\sin }^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} - \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} \cr & \Rightarrow \left( {\frac{1}{2} \times \frac{4}{3}} \right) + \left( {\frac{1}{4} \times \frac{2}{1}} \right) - \left( {\frac{1}{3} \times 1} \right) - \left( {\frac{1}{4} \times \frac{1}{3}} \right) \cr & \Rightarrow \frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{{12}} \cr & \Rightarrow \frac{1}{3} + \frac{1}{2} - \frac{1}{{12}} \cr & \Rightarrow \frac{{4 + 6 - 1}}{{12}} \cr & \Rightarrow \frac{9}{{12}} \cr & \Rightarrow \frac{3}{4} \cr} $$