Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \sec \theta + \frac{1}{{\cos \theta }} = 2 \cr & \sec \theta + \sec \theta = 2 \cr & \sec \theta = 1 \cr & {\sec ^{55}}\theta + \frac{1}{{{{\sec }^{55}}\theta }} \cr & = {\left( 1 \right)^{55}} + \frac{1}{{{{\left( 1 \right)}^{55}}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & \frac{{\cos A}}{{1 - \tan A}} + \frac{{\sin A}}{{1 - \cot A}} - \sin A \cr & = \frac{{\cos A}}{{1 - \frac{{\sin A}}{{\cos A}}}} + \frac{{\sin A}}{{1 - \frac{{\cos A}}{{\sin A}}}} - \sin A \cr & = \frac{{{{\cos }^2}A}}{{\cos A - \sin A}} + \frac{{{{\sin }^2}A}}{{\sin A - \cos A}} - \sin A \cr & = \frac{{{{\cos }^2}A}}{{\cos A - \sin A}} - \frac{{{{\sin }^2}A}}{{\cos A - \sin A}} - \sin A \cr & = \frac{{{{\cos }^2}A - {{\sin }^2}A}}{{\cos A - \sin A}} - \sin A \cr & = \frac{{{{\cos }^2}A - {{\sin }^2}A - \sin A\cos A + {{\sin }^2}A}}{{\cos A - \sin A}} \cr & = \frac{{{{\cos }^2}A - \sin A\cos A}}{{\cos A - \sin A}} \cr & = \frac{{\cos A\left( {\cos A - \sin A} \right)}}{{\cos A - \sin A}} \cr & = \boxed{\cos A} \cr & \cr & {\bf{Alternative: - }} \cr & {\text{You can put any }}A{\text{ value and satisfied equation,}} \cr & {\text{let }}A = {135^ \circ } \cr & \frac{{\cos {{135}^ \circ }}}{{1 - \tan {{135}^ \circ }}} + \frac{{\sin {{135}^ \circ }}}{{1 - \cot {{135}^ \circ }}} - \sin {135^ \circ } \cr & = \frac{{ - \frac{1}{{\sqrt 2 }}}}{2} + \frac{{ - \frac{1}{{\sqrt 2 }}}}{2} - \frac{1}{{\sqrt 2 }} \cr & = - \frac{1}{{\sqrt 2 }} \cr & {\text{Put in option (D) }}\cos {135^ \circ } = - \frac{1}{{\sqrt 2 }}{\text{ satisfied}} \cr & \therefore {\text{Option D is correct}}{\text{.}} \cr} $$

$$\eqalign{ & \sin A = \frac{5}{{13}} = \frac{P}{H};\,B = 12 \cr & \left[ {5,\,12,\,13\,{\text{are triplet}}} \right] \cr & 7\cot B = 24 \cr & \cot B = \frac{{24}}{7} = \frac{B}{P};\,H = 25 \cr & \left[ {7,\,24,\,25\,{\text{are triplet}}} \right] \cr & \Rightarrow \left( {\sec A\cos B} \right)\left( {{\text{cosec}}\,B\tan A} \right) \cr & = \left( {\frac{{13}}{{12}} \times \frac{{24}}{{25}}} \right)\left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{26}}{{25}} \times \left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{65}}{{42}} \cr} $$

$$\eqalign{ & \frac{{1 + \cos \theta - {{\sin }^2}\theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } }}{{\tan \theta + \cot \theta }} \cr & \Rightarrow \frac{{{{\cos }^2}\theta - \cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {\frac{1}{{{{\cos }^2}\theta }} + \frac{1}{{{{\sin }^2}\theta }}} }}{{\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}}} \cr & \Rightarrow \frac{{\cos \theta \left( {\cos \theta + 1} \right)}}{{\sin \theta \left( {1 + \cos \theta } \right)}} \times \frac{{\sqrt {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} }}{{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \frac{{\cos \theta }}{{\sin \theta }} \times \frac{{\sqrt {\frac{1}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} }}{{\frac{1}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \cot \theta \times \frac{{\frac{1}{{\cos \theta \sin \theta }}}}{{\frac{1}{{\cos \theta \sin \theta }}}} \cr & \Rightarrow \cot \theta \cr} $$

$$\eqalign{ & \frac{{\sin {{23}^ \circ }\cos {{67}^ \circ } + \sec {{52}^ \circ }\sin {{38}^ \circ } + \cos {{23}^ \circ }\sin {{67}^ \circ } + {\text{cosec}}\,{{52}^ \circ }\cos {{38}^ \circ }}}{{{\text{cose}}{{\text{c}}^2}{{20}^ \circ } - {{\tan }^2}{{70}^ \circ }}} \cr & = \frac{{{{\sin }^2}{{23}^ \circ } + \frac{1}{{\cos {{52}^ \circ }}} \times \cos {{52}^ \circ } + {{\cos }^2}{{23}^ \circ } + \frac{1}{{\sin {{52}^ \circ }}} \times \sin {{52}^ \circ }}}{{{\text{cose}}{{\text{c}}^2}{{20}^ \circ } - {{\cot }^2}{{20}^ \circ }}} \cr & = \frac{{{{\sin }^2}{{23}^ \circ } + 1 + {{\cos }^2}{{23}^ \circ } + 1}}{1}\,\,\,\,\,\left[ {\therefore \,{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right] \cr & = 1 + 1 + 1 \cr & = 3 \cr} $$

$$\eqalign{ & {\left\{ {\left( {\frac{{\sec \theta - 1}}{{\sec \theta + 1}}} \right)} \right\}^n} = {\text{cosec}}\,\theta - \cot \theta \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{So}},\,{\left\{ {\left( {\frac{{\sec {{45}^ \circ } - 1}}{{\sec {{45}^ \circ } + 1}}} \right)} \right\}^n} = {\text{cosec}}\,{45^ \circ } - \cot {45^ \circ } \cr & \Rightarrow {\left\{ {\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \right)} \right\}^n} = \sqrt 2 - 1 \cr & {\text{Rationalize internally on left side, we get}} \cr & \Rightarrow {\left\{ {{{\left( {\sqrt 2 - 1} \right)}^2}} \right\}^n} = \sqrt 2 - 1 \cr & {\text{To equate }}n{\text{ should be }}\frac{1}{2} \cr & {\text{So, }}n = 0.5 \cr} $$

$$\eqalign{ & \cos \theta = \frac{{12}}{{13}} = \frac{B}{H} \cr & P = \sqrt {{{13}^2} - {{12}^2}} = 5\,{\text{cm}} \cr & \frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}} \cr & = \frac{{\frac{P}{H}\left( {1 - \frac{P}{B}} \right)}}{{\frac{P}{B}\left( {1 + \frac{H}{P}} \right)}} \cr & = \frac{{\frac{5}{{13}}\left( {1 - \frac{5}{{12}}} \right)}}{{\frac{5}{{12}}\left( {1 + \frac{{13}}{5}} \right)}} \cr & = \frac{{\frac{5}{{13}} \times \frac{7}{{12}}}}{{\frac{5}{{12}} \times \frac{{18}}{5}}} \cr & = \frac{{\frac{{35}}{{156}}}}{{\frac{3}{2}}} \cr & = \frac{{35 \times 2}}{{156 \times 3}} \cr & = \frac{{35}}{{234}} \cr} $$

$$\eqalign{ & 2{\cos ^2}\theta + 3\sin \theta = 3 \cr & {\text{Let }}\theta = {30^ \circ } \cr & 2{\cos ^2}{30^ \circ } + 3\sin {30^ \circ } = 3 \cr & 2 \times \frac{3}{4} + 3 \times \frac{1}{2} = 3 \cr & 3 = 3 \cr & {\sin ^2}2\theta + {\cos ^2}\theta + {\tan ^2}2\theta + {\text{cose}}{{\text{c}}^2}2\theta \cr & = {\sin ^2}{60^ \circ } + {\cos ^2}{30^ \circ } + {\tan ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{60^ \circ } \cr & = \frac{3}{4} + \frac{3}{4} + 3 + \frac{4}{3} \cr & = \frac{3}{2} + 3 + \frac{4}{3} \cr & = \frac{{9 + 18 + 8}}{6} \cr & = \frac{{35}}{6} \cr} $$

$$\eqalign{ & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1 \cr & {\sin ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 1 - {\cos ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 0 = 2{\cos ^2}\theta - 3\cos \theta + 1 \cr & 2{\cos ^2}\theta - 3\cos \theta + 1 = 0 \cr & 2{\cos ^2}\theta - 2\cos \theta - \cos \theta + 1 = 0 \cr & 2\cos \theta \left( {\cos \theta - 1} \right) - 1\left( {\cos \theta - 1} \right) = 0 \cr & \left( {2\cos \theta - 1} \right)\left( {\cos \theta - 1} \right) = 0 \cr & \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \cos \theta = 1 = \cos {0^ \circ } \cr & \Rightarrow \frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }} \cr & = \frac{{{{\tan }^2}\frac{{{{60}^ \circ }}}{2} + {{\sin }^2}\frac{{{{60}^ \circ }}}{2}}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}}}{{\sqrt 3 + \frac{{\sqrt 3 }}{2}}} \cr & = \frac{{\frac{1}{3} + \frac{1}{4}}}{{\frac{{2\sqrt 3 + \sqrt 3 }}{2}}} \cr & = \frac{{\frac{{4 + 3}}{{12}}}}{{\frac{{3\sqrt 3 }}{2}}} \cr & = \frac{{7 \times 2}}{{12 \times 3\sqrt 3 }} \cr & = \frac{7}{{18\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{7\sqrt 3 }}{{54}} \cr} $$

$$\eqalign{ & \left( {{{\cos }^6}\theta + {{\sin }^6}\theta - 1} \right)\left( {{{\tan }^2}\theta + {{\cot }^2}\theta + 2} \right) \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \left( {{{\cos }^6}{{45}^ \circ } + {{\sin }^6}{{45}^ \circ } - 1} \right)\left( {{{\tan }^2}{{45}^ \circ } + {{\cot }^2}{{45}^ \circ } + 2} \right) \cr & = \left( {\frac{1}{8} + \frac{1}{8} - 1} \right)\left( {1 + 1 + 2} \right) \cr & = \frac{{ - 3}}{4} \times 4 \cr & = - 3 \cr} $$