Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta = 1\,......({\text{i}}) \cr & \because {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {{{\sec }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{{{{\sec }^2}\theta }} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{co}}{{\text{s}}^2}\theta \cr & \because {\text{ co}}{{\text{s}}^4}\theta + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{co}}{{\text{s}}^2}\theta } \right)^2} + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{ta}}{{\text{n}}^2}\theta } \right)^2} + {\text{ta}}{{\text{n}}^2}\theta \cr & = {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta \cr & = 1{\text{ from equation }}\left( {\text{i}} \right) \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr & \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr & \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$

Solution:

Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution:

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

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:

 $$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{{81}^ \circ }} \right)}}{{\sin {{56}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }{\text{cosec}}\,{{65}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{\cos {{34}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }\sec {{25}^ \circ }}} \cr & = \frac{{2\cos {9^ \circ } \times 2\sec {9^ \circ }}}{{1 + 1}} \cr & = \frac{{4 \times \cos {9^ \circ }\sec {9^ \circ }}}{2} \cr & = 2 \cr} $$

Solution:

Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution:

 $$\eqalign{ & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr & 4x - {50^ \circ } = {40^ \circ } + x \cr & 3x = {90^ \circ } \cr & x = {30^ \circ } \cr} $$

Solution:

 $$\eqalign{ & \sec A = \frac{{17 \to H}}{{8 \to B}},\,A {90^ \circ } \cr & {L^2} = {H^2} - {B^2} \cr & {L^2} = 289 - 64 \cr & {L^2} = 225 \cr & L = 15 \cr & {\text{Then}}, \cr & \frac{{34\sin A + 15\cot A}}{{68\cos A - 16\tan A}} \cr & = \frac{{34 \times \frac{{15}}{{17}} + 15 \times \frac{8}{{15}}}}{{68 \times \frac{8}{{17}} - 16 \times \frac{{15}}{8}}} \cr & = \frac{{30 + 8}}{{32 - 30}} \cr & = \frac{{38}}{2} \cr & = 19 \cr} $$