Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{Given 4}}{\sin ^2}\theta = 3 \cr & {\sin ^2}\theta = \frac{3}{4} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \sin \theta = {\text{sin }}{60^ \circ } \cr & \theta = {60^ \circ } \cr & \because \tan \theta - \cot \frac{\theta }{2} \cr & = \tan {60^ \circ } - \cot \frac{{{{60}^ \circ }}}{2} \cr & = \tan {60^ \circ } - \cot {30^ \circ } \cr & = \sqrt 3 - \sqrt 3 \cr & = 0 \cr} $$

sin31° = $$\frac{x}{y}{\text{,}}$$
∴ sec31° - sin59°
$$\eqalign{ & = \frac{y}{{\sqrt {{y^2} - {x^2}} }} - \frac{{\sqrt {{y^2} - {x^2}} }}{y} \cr & = \frac{{{y^2} - ({y^2} - {x^2})}}{{y\sqrt {{y^2} - {x^2}} }} \cr & = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {\cot {{73}^ \circ }{{\cos }^2}{{22}^ \circ } + \frac{1}{{\cot {{17}^ \circ }se{c^2}{{68}^ \circ }}}} \right)$$
$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left[ {\cot \left( {{{90}^ \circ } - {{17}^ \circ }} \right){{\cos }^2}\left( {{{90}^ \circ } - {{68}^ \circ }} \right) + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right]$$
$$\eqalign{ & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {tan{{17}^ \circ }si{n^2}{{68}^ \circ } + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\tan {17^ \circ }\left( {si{n^2}{{68}^ \circ } + {\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow 1\left( 1 \right) \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \sin \theta - \cos \theta = 0 \cr & \sin \theta = \cos \theta \cr & \theta = {45^ \circ } \cr & {\text{Then,}} \cr & sec\theta + {\text{cosec}}\theta \cr & = \sqrt 2 + \sqrt 2 \cr & = 2\sqrt 2 \cr} $$

$$\eqalign{ & \alpha + \beta = {90^ \circ } \cr & {\text{and }}\alpha :\beta = 2:1 \cr & 2x + x = {90^ \circ } \cr & x = {30^ \circ } \cr & {\text{ }}\alpha = {60^ \circ } \cr & \beta = {30^ \circ } \cr & \frac{{{\text{cos }}\alpha }}{{\cos \beta }} \cr & = \frac{{\cos {{60}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \cr & = \frac{1}{{\sqrt 3 }} \cr & = 1:\sqrt 3 \cr} $$

$$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$

$${\text{ta}}{{\text{n}}^2}x + {\text{co}}{{\text{t}}^2}x$$
We know that the value of tan²x and cot²x is minimum at 45°
$$\eqalign{ & \therefore {\text{ta}}{{\text{n}}^2}{45^ \circ } + {\text{co}}{{\text{t}}^2}{45^ \circ } \cr & = 1 + 1 \cr & = 2 \cr} $$

Given, cos27° = x

tan63° = $$\frac{x}{{\sqrt {1 - {x^2}} }}$$

$$\eqalign{ & {\cos ^2}x + {\cos ^4}x = 1 \cr & \Rightarrow {\cos ^4}x = 1 - {\cos ^2}x \cr & \Rightarrow {\cos ^4}x = {\sin ^2}x \cr & \Rightarrow {\cos ^2}x{\cos ^2}x = {\sin ^2}x \cr & \Rightarrow {\text{co}}{{\text{s}}^2}x = {\text{ta}}{{\text{n}}^2}x\,\,\,....(i) \cr & \Rightarrow {\cos ^4}x = {\text{ta}}{{\text{n}}^4}x\,\,\,....(ii) \cr & ta{n^2}x + {\tan ^4}x \cr & {\text{co}}{{\text{s}}^2}x + {\text{co}}{{\text{s}}^4}x = 1 \cr & {\text{ta}}{{\text{n}}^2}x + {\text{ta}}{{\text{n}}^4}x = 1 \cr & \left( {{\text{From eq}}{\text{. (i) and (ii)}}} \right) \cr} $$

$$\eqalign{ & {\text{ tan}}\theta = \frac{4}{3} \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{3} \cr & \Rightarrow \frac{{3 \times 4 + 2 \times 3}}{{3 \times 4 - 2 \times 3}} \cr & \Rightarrow \frac{{18}}{6} \cr & \Rightarrow 3 \cr} $$