Practice MCQ Questions and Answer on Trigonometry

Solution:

 \[\begin{array}{l} \frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \cot \theta + {\rm{cosec}}\,\theta } \right]}}{{\left[ {\cot \theta + {\rm{cosec}}\,\theta + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }}} \right]}}{{\left[ {\frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }} + 1} \right]}}\\ \Rightarrow \frac{{\left[ {\sin \theta - \cos \theta + 1} \right]}}{{\left[ {\sin \theta + \cos \theta + 1} \right]}}\\ \Rightarrow \frac{{\left( {\sin \theta + 1} \right) - \cos \theta }}{{\left( {\sin \theta + 1} \right) + \cos \theta }}\\ \left[ \begin{array}{l} \therefore \sin \theta = 2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\\ \cos \theta = 1 - 2{\sin ^2}\frac{\theta }{2}\\ \cos \frac{\theta }{2} = 2{\cos ^2}\frac{\theta }{2} - 1 \end{array} \right]\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 - 1 + 2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 + 2{{\cos }^2}\frac{\theta }{2} - 1}}\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}{{2\cos \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}\\ \Rightarrow \tan \frac{\theta }{2} \end{array}\]

Solution:

 $$\eqalign{ & 3{\sin ^2}{30^ \circ } + \frac{3}{5}{\cos ^2}{60^ \circ } - 2{\sec ^2}{45^ \circ } \cr & = 3 \times \frac{1}{4} + \frac{3}{5} \times \frac{1}{4} - 2 \times 2 \cr & = \frac{3}{4} + \frac{1}{5} \times \frac{3}{4} - 4 \cr & = \frac{3}{4} + \frac{3}{{20}} - 4 \cr & = \frac{{15 + 3 - 80}}{{20}} \cr & = \frac{{ - 62}}{{10}} \cr & = \frac{{ - 31}}{{10}} \cr} $$

Solution:

 cosec(67° + θ) - sec(23° - θ) + cos15°cos35°cosec55°cos60°cosec75° = sec[90° - (67° + θ)] - sec(23° - θ) + sin75°sin55°cosec55°cos60°cosec75° = sec(23° - θ) - sec(23° - θ) + $$\frac{1}{2}$$ = $$\frac{1}{2}$$

Solution:

 $$\eqalign{ & {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr & = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr & = \frac{{16}}{3} \cr & = 5\frac{1}{3} \cr} $$

Solution:

 $$\eqalign{ & {\text{sin P}} + {\text{cosec P}} = 2 \cr & {\text{For P}} = {90^ \circ } \cr & \Rightarrow {\text{sin }}{90^ \circ } + {\text{cosec }}{90^ \circ } = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & \Rightarrow 2 = 2\left( {{\text{satisfy}}} \right) \cr & {\text{So, }}{\sin ^7}{\text{P}} + {\text{cose}}{{\text{c}}^7}{\text{P}} \cr & \Rightarrow {\sin ^7}{90^ \circ } + {\text{cose}}{{\text{c}}^7}{90^ \circ } \cr & \Rightarrow {1^7} + {1^7} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & \sin \theta - \cos \theta = \frac{7}{{13}} = \alpha \cr & {\text{When, }} \cr & ax + by = m\,......(i) \cr & bx - ay = n\,......(ii) \cr} $$ By adding these two equations after making square on both sides we get, $$\eqalign{ & \left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) = {m^2} + {n^2} \cr & {\text{In the same process}} \cr & \sin \theta \pm \cos \theta = a \cr & {\text{Then,}}\sin \theta \pm \cos \theta = \sqrt {2 - {a^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - {{\left( {\frac{7}{{13}}} \right)}^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - \left( {\frac{{49}}{{169}}} \right)} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {\frac{{289}}{{169}}} \cr & \Rightarrow \sin \theta + \cos \theta = \frac{{17}}{{13}} \cr} $$

Solution:

 $$\eqalign{ & {\text{2cos}}\theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{When,}} \cr & ax \mp by = m \cr & {\text{then, }}bx \mp ay = \sqrt {{a^2} + {b^2} - {m^2}} \cr & 2\cos \theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \theta + 2\sin \theta = \sqrt {4 + 1 - \frac{1}{2}} \cr & \Rightarrow \cos \theta + 2\sin \theta = \frac{3}{{\sqrt 2 }} \cr} $$

Solution:

 $$\eqalign{ & {\text{ }}\sqrt 3 \tan \theta = 3\sin \theta \cr & {\bf{Shortcut method:}} \cr & \Rightarrow {\text{ }}\sqrt 3 \frac{{\sin \theta }}{{\cos \theta }} = 3\sin \theta \cr & \Rightarrow \frac{{\sqrt 3 }}{{\cos \theta }} = 3 \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{3} \cr & {\text{then perpendicular}} = \sqrt 6 \cr} $$ $$\eqalign{ & \Rightarrow \left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow {\left( {\frac{P}{H}} \right)^2} - {\left( {\frac{B}{H}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{3}} \right)^2} \cr & \Rightarrow \frac{6}{9} - \frac{3}{9} \cr & \Rightarrow \frac{1}{3} \cr} $$

Solution:

 $$\eqalign{ & 4 - 2{\sin ^2}\theta - 5\cos \theta = 0 \cr & {\text{Let }}\theta = {60^ \circ } \cr & 4 - 2{\sin ^2}{60^ \circ } - 5\cos {60^ \circ } = 0 \cr & 4 - 2 \times \frac{3}{4} - 5 \times \frac{1}{2} = 0 \cr & 4 - 4 = 0 \cr & \sin \theta + \tan \theta = \sin {60^ \circ } + \tan {60^ \circ } \cr & = \frac{{\sqrt 3 }}{2} + \sqrt 3 \cr & = \frac{{3\sqrt 3 }}{2} \cr} $$

Solution:

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