Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$${\text{152}}\left( {\sin {{30}^ \circ }\, + \,2{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,3\sin {{30}^ \circ }\, + \,4{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,.....\, + \,17\sin {{30}^ \circ }\, + \,18{\text{co}}{{\text{s}}^2}{{45}^ \circ }} \right)$$
$$ = 152\left\{ {\frac{1}{2}\, + \,2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}\, + \,3\, \times \,\frac{1}{2}\, + \,.....\,\, + \,17 \times \frac{1}{2}\, + \,18{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right\}$$
$$\eqalign{ & = 152\left\{ {\frac{1}{2}\, + \,1\, + \,1\frac{1}{2} + \,.....\,8\frac{1}{2}\, + \,9} \right\} \cr & = {\text{This is in A}}{\text{.P}}{\text{. where}} \cr & a = \frac{1}{2},{\text{ }}d = \frac{1}{2},{\text{ }}n = 18 \cr & = 152\left\{ {\frac{{18}}{2}\left( {2\, \times \,\frac{1}{2}\, + \,\left( {18\, - \,1} \right)\frac{1}{2}} \right)} \right\} \cr & = 152\left\{ {\frac{{18}}{2}\left( {1\, + \,\frac{{17}}{2}} \right)} \right\} \cr & = 152 \times 9 \times \frac{{19}}{2} \cr & = 12996 \cr & = \sqrt {12996} \cr & = 114 \cr} $$

$$\eqalign{ & x\sin {45^ \circ } = y\operatorname{cosec} {30^ \circ } \cr & \Rightarrow \frac{x}{y} = \frac{{{\text{cosec 3}}{{\text{0}}^ \circ }}}{{{\text{sin }}{{45}^ \circ }}} \cr & \Rightarrow \frac{x}{y} = \frac{2}{{\frac{1}{{\sqrt 2 }}}} \cr & \Rightarrow \frac{x}{y} = \frac{{2\sqrt 2 }}{1} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {\left( {\frac{{2\sqrt 2 }}{1}} \right)^4} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = \frac{{64}}{1} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {4^3} \cr} $$

$$\eqalign{ & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr & \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr & \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr & \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & {\text{Put }}\beta = {45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr & {\text{ tan }}\alpha = \sqrt 3 \cr & \alpha = {60^ \circ } \cr & {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr & = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr & = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr & = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0 \cr} $$

$$\eqalign{ & {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr & {\text{Put }}\theta = {45^ \circ } \cr & 1 + 1 = 2\left( {{\text{matched}}} \right) \cr & {\text{So, }}\theta = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr & \Rightarrow {1^{100}} + {1^{100}} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr & {\bf{Shortcut method:}} \cr & {\text{Take, }}\theta = {0^ \circ } \cr & \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr & \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{tan}}\theta - \cot \theta = 0 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{tan }}{45^ \circ } - \cot {45^ \circ } = 0 \cr & 1 - 1 = 0 \cr & 0 - 0({\text{matched}}) \cr & So,\theta = {45^ \circ } \cr & \Rightarrow \sin \theta + \cos \theta \cr & \Rightarrow \sin {45^ \circ } + \cos {45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \sqrt 2 \cr} $$

$$\eqalign{ & x\sin {60^ \circ }.\tan {30^ \circ } = \sec {60^ \circ }.\cot {45^ \circ } \cr & {\text{Put values }} \cr & \Rightarrow x.\frac{{\sqrt 3 }}{2}.\frac{1}{{\sqrt 3 }} = 2.1 \cr & \Rightarrow \frac{x}{2} = 2 \cr & \Rightarrow x = 4 \cr} $$

$$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$

$${\text{ }}x\sin {60^ \circ }.\tan {30^ \circ } - \tan^2 {45^ \circ } = $$       $$\operatorname{cosec} {60^ \circ }.$$   $$\cot {30^ \circ } - $$   $${\sec ^2}{45^ \circ }$$
$$\eqalign{ & \Rightarrow x \times \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 3 }} - 1 = \frac{2}{{\sqrt 3 }} \times \sqrt 3 - {\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow \frac{x}{2} - 1 = 2 - 2 \cr & \Rightarrow \frac{x}{2} - 1 = 0 \cr & \Rightarrow \frac{x}{2} = 1 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ } + 4{\text{se}}{{\text{c}}^2}{{30}^ \circ } - {\text{ta}}{{\text{n}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {\text{co}}{{\text{s}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{2}} \right)}^2} + 4{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - 1}}{1} \cr & \left( {{{\sin }^2}{\text{A}} + {{\cos }^2}{\text{A}} = {\text{1}}} \right) \cr & \Rightarrow \frac{1}{4} + \frac{{4 \times 4}}{3} - 1 \cr & \Rightarrow \frac{1}{4} + \frac{{16}}{3} - 1 \cr & \Rightarrow \frac{{3 + 64 - 12}}{{12}} \cr & \Rightarrow \frac{{55}}{{12}} \cr} $$