Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

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

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

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

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

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

$$\eqalign{ & xsin\theta + y\cos \theta = 4 \cr & \underline {x\cos \theta - y\sin \theta = 2} {\text{ }}\,\,\,\,\left[ {{\text{Formula}}} \right] \cr & \left( {{x^2} + {y^2}} \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = {4^2} + {2^2} \cr & \Rightarrow {x^2} + {y^2} = {a^2} + {b^2} \cr & \Rightarrow \left( {{x^2} + {y^2}} \right)\left( 1 \right) = 16 + 4 \cr & \Rightarrow {x^2} + {y^2} = 20 \cr} $$

  $$\left[ {\frac{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.si}}{{\text{n}}^2}{\text{A}}\left( {{\text{sin A}} + {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)}} + \frac{{{\text{si}}{{\text{n}}^2}{\text{A}}{\text{.co}}{{\text{s}}^2}{\text{A}}\left( {{\text{sin A}} - {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$           $$\left[ {\frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} - \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}}} \right]$$
  $$ \Rightarrow \left[ {\frac{{{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}^2} + {{\left( {{\text{sin A}} - {\text{cos A}}} \right)}^2}}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$       $$\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} - {\text{co}}{{\text{s}}^2}{\text{ A}}} \right)$$
$$\eqalign{ & \Rightarrow 2\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} + {\text{co}}{{\text{s}}^2}{\text{ A}}} \right) \cr & \Rightarrow 2 \cr} $$

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

$$\eqalign{ & \frac{{{\text{sin A}}}}{{1 + \cos {\text{ A}}}}{\text{ + }}\frac{{{\text{sin A}}}}{{1 - \cos {\text{ A}}}} \cr & \Rightarrow \frac{{{\text{sin A}}\left( {1 - \cos {\text{ A}}} \right) + {\text{sin A}}\left( {1 + \cos {\text{ A}}} \right)}}{{\left( {1 + \cos {\text{ A}}} \right)\left( {1 - \cos {\text{ A}}} \right)}} \cr & \Rightarrow \frac{{{\text{sin A}} - {\text{sin A}}{\text{.cosA}} + {\text{sin A}} + {\text{sin A}}{\text{.cosA}}}}{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow \frac{{2{\text{sin A}}}}{{{{\sin }^2}{\text{A}}}} \cr & \Rightarrow 2{\text{ cosec A}} \cr} $$

$$\eqalign{ & r\sin \theta = 1 \cr & r\cos \theta = \sqrt 3 \cr & \Rightarrow \frac{{{\text{ }}rsin\theta }}{{{\text{ }}r\cos \theta }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow {\text{tan}}\theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \sqrt 3 {\text{tan}}\theta = 1 \cr & \left( {{\text{Add 1 both sides}}} \right) \cr & \Rightarrow \sqrt 3 {\text{tan}}\theta + 1 = 1 + 1 \cr & \Rightarrow \sqrt 3 {\text{tan}}\theta + 1 = 2 \cr} $$