Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1 \cr & = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{\left( {\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)^2} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

Solution:

 $$\eqalign{ & \tan {10^ \circ }.\tan {15^ \circ }.\tan {75^ \circ }.\tan {80^ \circ } \cr & \Rightarrow \left( {\tan {{10}^ \circ }.\tan {{80}^ \circ }} \right).\left( {\tan {{15}^ \circ }.\tan {{75}^ \circ }} \right) \cr & \Rightarrow 1 \times 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{Arc length}} = \frac{\theta }{{180}} \times \pi r \cr & \Rightarrow 44 = \frac{{18}}{{180}} \times \frac{{22}}{7} \times r \cr & \Rightarrow r = 140\,{\text{m}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{cot{{33}^ \circ } + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{\frac{1}{{\tan {{33}^ \circ }}} + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \frac{1}{{\tan {{53}^ \circ }}}}} \cr & = \frac{{1 + \tan {{53}^ \circ }.\tan {{33}^ \circ }}}{{\tan {{33}^ \circ }.\tan {{53}^ \circ } + 1}} \times \frac{{\tan {{53}^ \circ }}}{{\tan {{33}^ \circ }}} \cr & = \tan {53^ \circ }.cot{33^ \circ } \cr & = \cot {37^ \circ }.\tan {57^ \circ } \cr} $$

Solution:

 $$\eqalign{ & \frac{{2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)}}{{{{\cos }^4}\theta - {{\sin }^4}\theta - 2{{\cos }^2}\theta }} \cr & = \frac{{2\left( {1 - 3{{\sin }^2}\theta .{{\cos }^2}\theta } \right) - 3\left( {1 - 2{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) - 2{{\cos }^2}\theta }} \cr & = \frac{{2 - 6{{\sin }^2}\theta .{{\cos }^2}\theta - 3 + 6{{\sin }^2}\theta .{{\cos }^2}\theta }}{{ - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}} \cr & = \frac{{ - 1}}{{ - 1}} \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right) = {\text{B}} \times \tan \left( {\theta - {{60}^ \circ }} \right) \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {\theta - {{60}^ \circ }} \right)}}{{\tan \left( {\theta + {{150}^ \circ }} \right)}} \cr & {\text{Put }}\theta = {90^ \circ } \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right)}}{{\tan \left( {{{90}^ \circ } + {{150}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan \left( {{{180}^ \circ } + {{60}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{1}{3} \cr & {\text{then, }}\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - \frac{4}{2} \cr & \Rightarrow \frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - 2 \cr & \Rightarrow \frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}} = - \frac{1}{2} \cr & {\text{Put in option (i)}} \cr & - \frac{{\sin {{90}^ \circ }}}{2} = - \frac{1}{2} \cr & {\text{So, option (A) is correct }} \cr} $$

Solution:

 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)   $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$   $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$     $${\text{2cos5}}{{\text{9}}^ \circ }$$  $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$ $$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\left( {1 - \sin \theta + \cos \theta } \right)}^2}\left( {1 - \cos \theta } \right){{\sec }^3}\theta {\text{cose}}{{\text{c}}^2}\theta }}{{\left( {\sec \theta - \tan \theta } \right)\left( {\tan \theta + \cot \theta } \right)}} \cr & {\text{put }}\theta = {30^ \circ } \cr & = \frac{{{{\left( {1 - \sin {{30}^ \circ } + \cos {{30}^ \circ }} \right)}^2}\left( {1 - \cos {{30}^ \circ }} \right){{\sec }^3}{{30}^ \circ }{\text{cose}}{{\text{c}}^2}{{30}^ \circ }}}{{\left( {\sec {{30}^ \circ } - \tan {{30}^ \circ }} \right)\left( {\tan {{30}^ \circ } + \cot {{30}^ \circ }} \right)}} \cr & = \frac{{{{\left( {1 - \frac{1}{2} + \frac{{\sqrt 3 }}{2}} \right)}^2}\left( {1 - \frac{{\sqrt 3 }}{2}} \right){{\left( {\frac{2}{{\sqrt 3 }}} \right)}^3}{{\left( 2 \right)}^2}}}{{\left( {\frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }}} \right)\left( {\frac{1}{{\sqrt 3 }} + \sqrt 3 } \right)}} \cr & = \frac{{{{\left( {\frac{{1 + \sqrt 3 }}{2}} \right)}^2}\left( {\frac{{2 - \sqrt 3 }}{2}} \right){{\left( {\frac{2}{{\sqrt 3 }}} \right)}^3}{{\left( 2 \right)}^2}}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)\left( {\frac{{1 + 3}}{{\sqrt 3 }}} \right)}} \cr & = \frac{{\frac{{4 + 2\sqrt 3 }}{4} \times \frac{{2 - \sqrt 3 }}{2} \times \frac{8}{{3\sqrt 3 }} \times 4}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)\left( {\frac{4}{{\sqrt 3 }}} \right)}} \cr & = \frac{{\left[ {{2^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \times \frac{8}{{3\sqrt 3 }}}}{{\frac{4}{3}}} \cr & = \frac{{\frac{8}{{3\sqrt 3 }}}}{{\frac{4}{3}}} \cr & = \frac{2}{{\sqrt 3 }} \cr & {\text{Option A: }}2\tan {30^ \circ } = \frac{2}{{\sqrt 3 }} \cr & {\text{Option B: }}\cot {30^ \circ } = \sqrt 3 \cr & {\text{Option C: }}\sin {30^ \circ } = \frac{1}{2} \cr & {\text{Option D: }}2\cos {30^ \circ } = 2 \times \frac{{\sqrt 3 }}{2} = \sqrt 3 \cr & {\text{Only option A is answer}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{{\text{k}} + 1}}{{{\text{k}} - 1}} \cr & \frac{{\tan \theta }}{{\sin \theta }} = {\text{k}} \cr & {\text{k}} = \sec \theta \cr} $$

Solution:

 $$\eqalign{ & \tan \theta + \sin \theta = m \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta + 2{\text{ tan}}\theta .\sin \theta = {m^2}\,....(i) \cr & {\text{tan}}\theta - \sin \theta = n \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta - 2{\text{ tan}}\theta .\sin \theta = {n^2}\,....(ii) \cr & {\text{Substract from (i) and (ii)}} \cr & {m^2} - {n^2} = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta - {\text{ta}}{{\text{n}}^2}\theta - {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta \cr & {m^2} - {n^2} = 4{\text{tan}}\theta \sin \theta \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta {{\sin }^2}\theta } \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta - {{\sin }^2}\theta } \cr & = 4\sqrt {mn} \cr} $$