Practice MCQ Questions and Answer on Trigonometry

Solution:

 $$\eqalign{ & 5\sin \theta - 4\cos \theta = 0 \cr & 5\sin \theta = 4\cos \theta \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{5} \cr & \tan \theta = \frac{4}{5} \cr & \frac{{5\sin \theta - 2\cos \theta }}{{5\sin \theta + 3\cos \theta }} \cr & = \frac{{5\tan \theta - 2}}{{5\tan \theta + 3}} \cr & = \frac{{5 \times \frac{4}{5} - 2}}{{5 \times \frac{4}{5} + 2}} \cr & = \frac{2}{7} \cr} $$

Solution:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$

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:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$

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:

 $$\eqalign{ & \cos \theta = \frac{{12}}{{13}} = \frac{B}{H} \cr & P = \sqrt {{{13}^2} - {{12}^2}} = 5\,{\text{cm}} \cr & \frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}} \cr & = \frac{{\frac{P}{H}\left( {1 - \frac{P}{B}} \right)}}{{\frac{P}{B}\left( {1 + \frac{H}{P}} \right)}} \cr & = \frac{{\frac{5}{{13}}\left( {1 - \frac{5}{{12}}} \right)}}{{\frac{5}{{12}}\left( {1 + \frac{{13}}{5}} \right)}} \cr & = \frac{{\frac{5}{{13}} \times \frac{7}{{12}}}}{{\frac{5}{{12}} \times \frac{{18}}{5}}} \cr & = \frac{{\frac{{35}}{{156}}}}{{\frac{3}{2}}} \cr & = \frac{{35 \times 2}}{{156 \times 3}} \cr & = \frac{{35}}{{234}} \cr} $$

Solution:

 sec4θ = cosec(θ + 20°) sec4θ = sec[90° - (θ + 20°)] sec4θ = sec(70° - θ) 4θ = 70° - θ 5θ = 70° θ = 14°

Solution:

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

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:

 $$\eqalign{ & \left( {{{\cos }^6}\theta + {{\sin }^6}\theta - 1} \right)\left( {{{\tan }^2}\theta + {{\cot }^2}\theta + 2} \right) \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \left( {{{\cos }^6}{{45}^ \circ } + {{\sin }^6}{{45}^ \circ } - 1} \right)\left( {{{\tan }^2}{{45}^ \circ } + {{\cot }^2}{{45}^ \circ } + 2} \right) \cr & = \left( {\frac{1}{8} + \frac{1}{8} - 1} \right)\left( {1 + 1 + 2} \right) \cr & = \frac{{ - 3}}{4} \times 4 \cr & = - 3 \cr} $$