Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\text{8}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{12}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) \cr & {\text{Put }}\theta = {0^ \circ } \cr & = 8\left( {{{\sin }^6}{0^ \circ } + {{\cos }^6}{0^ \circ }} \right) - 12\left( {{{\sin }^4}{0^ \circ } + {{\cos }^4}{0^ \circ }} \right) \cr & = 8\left( {0 + 1} \right) - 12\left( {0 + 1} \right) \cr & = - 4 \cr} $$

Solution:

 $$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$

Solution:

 $$\eqalign{ & {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr & {\bf{Shortcut method:}} \cr & {\text{Take, }}\theta = {0^ \circ } \cr & \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr & \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr & \Rightarrow 2 \cr} $$

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:

 $$\eqalign{ & 5{\sin ^2}{60^ \circ } + 7{\sin ^2}{45^ \circ } + 8{\cos ^2}{45^ \circ } \cr & = 5{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 7{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 8{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{{15}}{4} + \frac{7}{2} + \frac{8}{2} \cr & = \frac{{15 + 14 + 16}}{4} \cr & = \frac{{45}}{4} \cr} $$

Solution:

 $$\eqalign{ & \tan \theta = \frac{3}{4} \cr & {\text{cos2}}\theta = \frac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} \cr & {\text{cos2}}\theta = \frac{7}{{25}} \cr} $$

Solution:

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

Solution:

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

Solution:

 $$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}}{\text{ + }}\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ }}}{{{{\sin }^2}{{45}^ \circ }}} - \frac{{{\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{45}^ \circ }}} - \frac{{{{\sin }^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} - \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} \cr & \Rightarrow \left( {\frac{1}{2} \times \frac{4}{3}} \right) + \left( {\frac{1}{4} \times \frac{2}{1}} \right) - \left( {\frac{1}{3} \times 1} \right) - \left( {\frac{1}{4} \times \frac{1}{3}} \right) \cr & \Rightarrow \frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{{12}} \cr & \Rightarrow \frac{1}{3} + \frac{1}{2} - \frac{1}{{12}} \cr & \Rightarrow \frac{{4 + 6 - 1}}{{12}} \cr & \Rightarrow \frac{9}{{12}} \cr & \Rightarrow \frac{3}{4} \cr} $$

Solution:

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