Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\text{cos}}\theta + \sin \theta = \sqrt 2 \cos \theta \cr & {\text{Squaring both sides}} \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta + 2\cos \theta .\sin \theta = 2{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2{\text{cos}}\theta .{\text{sin}}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\sqrt 2 \cos \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \cos \theta - \sin \theta = \frac{{2\sin \theta .\cos \theta }}{{\sqrt 2 \cos \theta }} \cr & \Rightarrow \sqrt 2 \sin \theta \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}\sqrt 2 \cos \theta = \alpha \cr & \therefore \cos \theta \pm \sin \theta = a \cr & \cos \theta \pm \sin \theta = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - 2{\text{co}}{{\text{s}}^2}\theta } \cr & = \sqrt {2\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = \sqrt {2{{\sin }^2}\theta } \cr & = \sqrt 2 \sin \theta \cr} $$

Solution:

 $$\cos \theta = \frac{{15 \to {\text{Base}}}}{{17 \to {\text{Hypo}}}}$$ $$\eqalign{ & {\text{Perpendicular = 8}} \cr & \Rightarrow {\text{cot}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{tan}}\theta = \frac{8}{{15}}\left[ {\therefore \tan \theta = \frac{{\text{P}}}{{\text{B}}}} \right] \cr} $$

Solution:

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

Solution:

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

Solution:

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

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:

 $$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$

Solution:

 $$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$

Solution:

 $$\eqalign{ & {\text{sin}}\left( {{{90}^ \circ } - \theta } \right) + {\text{cos}}\theta = \sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{cos}}\theta + {\text{cos}}\theta = \sqrt 2 \sin \theta \cr & \Rightarrow \frac{{2\cos \theta }}{{\sin \theta }} = \sqrt 2 \cr & \Rightarrow \cot \theta = \frac{{1 \to {\text{B}}}}{{\sqrt 2 \to {\text{P}}}} \cr & {\text{So, H}} \to \text{alignment} \cr & \therefore {\text{cosec}}\theta = \frac{{\text{H}}}{{\text{P}}} = \frac{{\sqrt 3 }}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{3}{2}} \cr} $$

Solution:

 cosec(65° + θ) - sec(25° - θ) + tan220° - cosec270° = sec(25° - θ) - sec(25° - θ) + tan220° - sec220° = tan220° - sec220° = -1