Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

Shortcut method :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$     $$\left( {1 + {{\cot }^2}\beta } \right)$$  $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & {\text{Put }} \alpha = \beta = {45^ \circ } \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{45^ \circ }.{\sin ^2}{45^ \circ }.{\text{cose}}{{\text{c}}^2}{45^ \circ }.{\operatorname{sce} ^2}{45^ \circ } \cr & \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2 \cr & \Rightarrow 1 \cr} $$

Alternate :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$     $$\left( {1 + {{\cot }^2}\beta } \right)$$  $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \times \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - \beta } \right).{\sin ^2}\alpha .{\text{cose}}{{\text{c}}^2}\beta .{\text{se}}{{\text{c}}^2}\left( {{{90}^ \circ } - \alpha } \right) \cr & \Rightarrow {\sin ^2}\beta .{\text{cose}}{{\text{c}}^2}\beta .{\sin ^2}\alpha . {\text{cose}}{{\text{c}} ^2}\alpha \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & {\text{ tan }}{9^ \circ } = \frac{p}{q} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{{{\operatorname{cosec} }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}{{81}^ \circ }}} \times {\sin ^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\left( {{{90}^ \circ } - {9^ \circ }} \right) \cr & \Rightarrow {\text{co}}{{\text{t}}^2}{9^ \circ } \cr & \Rightarrow \frac{{{q^2}}}{{{p^2}}} \cr} $$

The value of,
cos24° + cos55° + cos125° + cos204° + cos300°
We know that, cos(180° $$ \pm $$ θ) = -cosθ
⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°)
⇒ cos24° + cos55° - cos55° - cos24° + cos60°
⇒ cos60°
⇒ $$\frac{1}{2}$$

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$

$$\eqalign{ & {\text{tan}}\left( {2\theta + {{45}^ \circ }} \right) = \cot 3\theta \cr & \left[ {{\text{If tan A}} = {\text{cot B}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \Rightarrow \left( {2\theta + {{45}^ \circ }} \right) + 3\theta = {90^ \circ } \cr & \Rightarrow 5\theta + {45^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{5} \cr & \Rightarrow \theta = {9^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{\text{A}} - {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} - {\text{B}} = {\text{3}}{0^ \circ }} \right) \cr & {\text{cos}}\left( {{\text{A}} + {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} + {\text{B}} = {{60}^ \circ }} \right) \cr & {\text{Adding}}\,{\text{both}}\,{\text{side}} \cr & \Rightarrow \left( {{\text{A}} - {\text{B}}} \right) + \left( {{\text{A + B}}} \right) = {30^ \circ } + {60^ \circ } \cr & \Rightarrow 2{\text{A}} = {90^ \circ } \cr & \Rightarrow {\text{A}} = {45^ \circ } \cr & \because {\text{A}} - {\text{B}} = {30^ \circ } \cr & {\text{B}} = {\text{A}} - {30^ \circ } \cr & \Rightarrow {45^ \circ } - {30^ \circ } \cr & \Rightarrow {15^ \circ } \cr & \Leftrightarrow \frac{{15 \times \pi }}{{180}} = \frac{\pi }{{12}}({\text{radian}}) \cr} $$

$$\eqalign{ & {\sin ^2}\theta - 3\sin \theta + 2 = 0 \cr & \Rightarrow {\sin ^2}\theta - 2{\text{sin }}\theta - \sin \theta + 2 = 0 \cr & \Rightarrow \sin \theta \left( {\sin \theta - 2} \right) - 1\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \left( {\sin \theta - 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \left[ {\because \sin \theta \ne 2} \right]{\text{Put value of }} \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow {\text{sin }}\theta = \sin {90^ \circ } \cr & \Rightarrow \theta = {90^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }}\theta = {90^ \circ } \cr & \left[ {{\text{Take help from the options}}} \right] \cr & \Rightarrow {\sin ^2}\theta - 3\sin {\text{ }}\theta + 2 = 0 \cr & \Rightarrow {\sin ^2}{90^ \circ } - 3{\text{sin }}{90^ \circ } + 2 = 0 \cr & \Rightarrow 1 - 3 \times 1 + 2 = 0 \cr & \left[ {\sin {{90}^ \circ } = 1} \right] \cr & \Rightarrow 0 = 0\left[ {{\text{matched}}} \right] \cr & {\text{So, this is answer}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{2}}\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = 1 + \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = \frac{3}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{4} \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\theta = \frac{4}{3} \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}\theta = \frac{4}{3} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{4}{3} - 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{3} \cr & \Rightarrow \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \cot \theta = \sqrt 3 \cr & \cr & {\bf{Alternate:}} \cr & {\text{2}}\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {{\text{1}} - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = 1 + \frac{1}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{2} \times \frac{1}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{4} \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{2}\left[ {\cos {{30}^ \circ } = \frac{{\sqrt 3 }}{2}} \right] \cr & \Rightarrow \theta = {30^ \circ } \cr & {\text{Hence, }} \cr & {\text{cot}}\theta = {\text{cot3}}{0^ \circ } = \sqrt 3 \cr} $$

$$\eqalign{ & \sin {\text{ }}\alpha + \cos \beta = 2 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Leftrightarrow {\text{sin }}{90^ \circ } + {\text{cos }}{0^ \circ } = 2 \cr & \Leftrightarrow 1 + 1 = 2 \cr & \Leftrightarrow 2 = 2\left[ {{\text{Matched}}} \right] \cr & {\text{So, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }{\text{ }} \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2 \times {{90}^ \circ } + {0^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{{{180}^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{60^ \circ } = \cos {30^ \circ } = \frac{{\sqrt 3 }}{2} \cr & {\text{Take cos}}\frac{\alpha }{3} = \cos \frac{{{{90}^ \circ }}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{cos 3}}{0^ \circ } \cr & {\text{So, this is answer }}{\text{.}} \cr} $$

$$\eqalign{ & 2\sin \left( {\frac{{\pi x}}{2}} \right) = {x^2} + \frac{1}{{{x^2}}} \cr & {\text{Let }}x = 1 \cr & \Rightarrow 2\sin {90^ \circ } = {1^2} + \frac{1}{{{1^2}}} \cr & \Rightarrow 2 \times 1 = 1 + 1 \cr & 2 = 2\left( {{\text{Matched}}} \right) \cr & {\text{So, }}x = 1 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) \cr & \Rightarrow 1 - \frac{1}{1} \cr & \Rightarrow 0 \cr} $$