Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

Answer: Option B

According to the question,
  $$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$       $${\sin ^2}{67^ \circ } - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$       $${\sin ^2}{23^ \circ } - $$   $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$     $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$   $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$\eqalign{ & \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr & \Rightarrow se{c^2}{23^ \circ } \cr} $$

Answer: Option D

$$\eqalign{ & \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr & \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr & \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$

Answer: Option B

$$\eqalign{ & {\bf{Shortcut \,\, method:}} \cr & = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr & \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$
  $$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$     $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$
$$\eqalign{ & = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr & = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr & = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr & = 4 - 2 \cr & = 2 \cr} $$

Answer: Option A

$$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

Answer: Option D

$$\eqalign{ & \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr & \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow 3\left( 1 \right) \cr & \Rightarrow 3 \cr} $$