Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr & 4x - {50^ \circ } = {40^ \circ } + x \cr & 3x = {90^ \circ } \cr & x = {30^ \circ } \cr} $$

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

$$\eqalign{ & {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr & {\text{sin A}} = {\text{cos B}} \cr & {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr & {\text{A}} = {90^ \circ } - {\text{B}} \cr & {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \angle {\text{C}} = {90^ \circ } \cr & {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$

$$\eqalign{ & \sin \theta \times \cos \theta = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & 2\sin \theta \times \cos \theta = 1 \cr & \sin 2\theta = 1 \cr & \sin 2\theta = \sin {90^ \circ } \cr & 2\theta = {90^ \circ } \cr & \theta = {45^ \circ } \cr & {\text{So,}}\sin \theta - {\text{cos}}\theta \cr & = \sin {45^ \circ } - \cos {45^ \circ } \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0{\text{ }} \cr} $$

$$\eqalign{ & = {\cos ^2}{20^ \circ } + {\cos ^2}{70^ \circ } \cr & = {\cos ^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\cos ^2}{70^ \circ } \cr & = 1 \cr} $$

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

$$\eqalign{ & \sec 15\theta = \operatorname{cosec} 15\theta \cr & \Rightarrow \frac{1}{{\cos {{15} }\theta }} = \frac{1}{{\sin 15\theta }} \cr & \Rightarrow \frac{{\sin 15\theta }}{{{\text{cos15}}\theta }} = 1 \cr & \Rightarrow {\text{tan15}}\theta = 1 \cr & \Rightarrow {\text{tan15}}\theta = {\text{tan}}{45^ \circ } \cr & \Rightarrow 15\theta = {45^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{{15}} \cr & \Rightarrow \theta = {3^ \circ } \cr} $$

$$\eqalign{ & {\text{tan}}\theta = {\text{tan}}{30^ \circ }.{\text{tan}}{60^ \circ } \cr & {\text{tan}}\theta = \frac{1}{{\sqrt 3 }}.\sqrt 3 \cr & {\text{tan}}\theta = 1 \cr & {\text{tan}}\theta = {\text{tan}}{45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore 2\theta = {90^ \circ } \cr} $$

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & 4{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{4} \cr & {\text{sin }}\theta = \frac{1}{2} \cr & \sin \theta = \sin {30^ \circ } \cr & \theta = {30^ \circ } \cr & \sec \theta + \operatorname{cosec} \theta \cr & = \sec {30^ \circ } + \operatorname{cosec} {30^ \circ } \cr & = \frac{2}{{\sqrt 3 }} + 2 \cr} $$

$$\eqalign{ & \left( {\frac{{\sin \theta + \sin \phi }}{{\cos \theta + \cos \phi }} + \frac{{\cos \theta \cos \phi }}{{\sin\theta - \sin\phi }}} \right) \cr & {\text{Put }}\theta = {90^ \circ } \cr & \phi = {0^ \circ } \cr & \therefore \left( {\frac{{\sin90^ \circ + \sin0^ \circ }}{{\cos90^ \circ + \cos0^ \circ }} + \frac{{\cos90^ \circ - \cos0^ \circ }}{{\sin90^ \circ - \sin0^ \circ }}} \right) \cr & \Rightarrow \left( {\frac{{1 + 0}}{{0 + 1}} + \frac{{0 - 1}}{{1 - 0}}} \right) \cr & \Rightarrow \left( {1 - 1} \right) \cr & \Rightarrow 0 \cr} $$