Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{sin }}\alpha .\sec \left( {{{30}^ \circ } + \alpha } \right) = 1 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\alpha {\text{ value between }}{{\text{0}}^ \circ }{\text{ to 6}}{{\text{0}}^ \circ } \cr & {\text{If }}\alpha = {30^ \circ } \cr & \Rightarrow {\text{sin 3}}{0^ \circ }.\sec \left( {{{30}^ \circ } + {{30}^ \circ }} \right) = 1 \cr & \Rightarrow {\text{sin 3}}{0^ \circ }.\sec {60^ \circ } = 1 \cr & \Rightarrow \frac{1}{2} \times 2 \cr & \Rightarrow 1 \cr & 1 = 1\left( {{\text{Satisfy}}} \right) \cr & {\text{So}},\alpha = {30^ \circ } \cr & = \sin \alpha + \cos 2\alpha \cr & = \sin {30^ \circ } + \cos 2 \times {30^ \circ } \cr & = \sin {30^ \circ } + \cos {60^ \circ } \cr & = \frac{1}{2} + \frac{1}{2} \cr & = 1 \cr } $$

$$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$
  $$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$       $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$   $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$
$$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$

cotθ. tan(90° - θ) - sec(90° - θ)cosecθ + (sin²25° + sin²65°) + $$\sqrt 3 $$ (tan5°. tan15°. tan30°. tan75°. tan85°)
= cotθ. cotθ - cosecθ. cosecθ + (sin²25° + cos²25° ) + $$\sqrt 3 $$ [(tan5°. tan85°) . (tan15°. tan75°). tan30°]
  $$ = \left( {{\text{co}}{{\text{t}}^2}\theta - {\text{cose}}{{\text{c}}^2}\theta } \right) + \left( 1 \right) + \sqrt 3 \left( {1.1.\frac{1}{{\sqrt 3 }}} \right)$$         $$\left[ {{\text{tan A}}{\text{.tan B}} = {\text{1, If A}} + {\text{B}} = {{90}^ \circ }} \right]$$
$$\eqalign{ & = \left( { - 1} \right) + \left( 1 \right) + \sqrt 3 \times \frac{1}{{\sqrt 3 }} \cr & = - 1 + 1 + 1 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{cos}}\theta .{\text{cosec2}}{{\text{3}}^ \circ } = {\text{1}} \cr & {\text{If cosA}}.{\text{cosecB}} = {\text{1}} \cr & {\text{Then, A + B}} = {90^ \circ } \cr & {\text{So,}} \cr & \theta {\text{ + 2}}{{\text{3}}^ \circ } = {90^ \circ } \cr & \theta = {67^ \circ } \cr} $$

$$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} = \pi = {\text{18}}{0^ \circ }{\text{ }} \cr & \Rightarrow \frac{{{\text{A + B}}}}{2} = \frac{{{{180}^ \circ }}}{2} - \frac{{\text{C}}}{2} \cr & \Rightarrow {\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) \cr & \Rightarrow {\text{sin }}\left( {\frac{\pi }{2} - \frac{{\text{C}}}{2}} \right) \cr & \Rightarrow {\text{cos}}\frac{{\text{C}}}{2} \cr & \cr & {\bf{Similarly:}} \cr & {\text{cos }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2} \cr & {\text{cot }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2} \cr & {\text{tan }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \cot \frac{{\text{C}}}{2} \cr & {\text{So, option C is incorrect}}{\text{.}} \cr} $$

$$\eqalign{ & \tan 2\theta .\tan 3\theta = 1 \cr & \left( {2\theta + 3\theta } \right) = {90^ \circ } \cr & 5\theta = {90^ \circ } \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & \left[ {{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {{\text{2co}}{{\text{s}}^2}\frac{{5\theta }}{2} - 1} \right) \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}\frac{{{{90}^ \circ }}}{2} - 1 \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}{45^ \circ } - 1 \cr & \Rightarrow \frac{{\text{2}}}{2} - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \tan 7\theta .\tan 2\theta = 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \left( {7\theta + 2\theta } \right) = {90^ \circ } \cr & 9\theta = {90^ \circ } \cr & \theta = {10^ \circ } \cr & \Rightarrow \tan 3\theta \cr & \Rightarrow \tan {30^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{{60}^ \circ } - \theta } \right) = {\text{cos}}\left( {\psi - {{30}^ \circ }} \right) \cr & \Rightarrow \left( {{{60}^ \circ } - \theta } \right) + \left( {\psi - {{30}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}\,{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {\psi - \theta } \right) = {90^ \circ } - {30^ \circ } \cr & \Rightarrow \left( {\psi - \theta } \right) = {60^ \circ } \cr & \Rightarrow \tan \left( {\psi - \theta } \right) = \tan {60^ \circ } \cr & \Rightarrow \tan {60^ \circ } = \sqrt 3 \cr} $$

3cos80°.cosec10° + 2cos59°.cosec31°
[If A + B = 90° then, cosA.cosecB = 1]
(3 × 1 + 2 × 1 = 5)
(the value of sinθ is always between -1 to +1)
  $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$
  $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$     $${\text{2cos5}}{{\text{9}}^ \circ }$$  $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$
$$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

  $$\frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos 2}}{{\text{2}}^ \circ }}}$$   $$ - $$ $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan {{75}^ \circ }}}$$   $$ - $$ $$\frac{{3\tan {{45}^ \circ }.\tan {{20}^ \circ }.\tan {{40}^ \circ }.\tan {{50}^ \circ }.\tan {{70}^ \circ }}}{5}$$
  $$ \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos }}\left( {{{90}^ \circ } - {{68}^ \circ }} \right)}} - $$     $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan \left( {{{90}^ \circ } - {{15}^ \circ }} \right)}} - $$     $$\frac{{3.1\left( {\tan {{20}^ \circ }.\tan {{70}^ \circ }} \right).\left( {\tan {{40}^ \circ }.\tan {{50}^ \circ }} \right)}}{5}$$
$$\eqalign{ & \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{sin 6}}{{\text{8}}^ \circ }}} - \frac{{2{\text{cot 1}}{5^ \circ }}}{{5\cot {{15}^ \circ }}} - \frac{{3 \times 1 \times 1 \times 1}}{5} \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1 then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2 - \frac{2}{5} - \frac{3}{5} \cr & \Rightarrow 1 \cr} $$