Practice MCQ Questions and Answer on Trigonometry

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:

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

Solution:

 sin238° + sin252° + sin230° - tan245° = sin238° + cos238° + $${\left( {\frac{1}{2}} \right)^2}$$ - 1   [Here sin2θ + cos2θ = 1] = 1 + $$\frac{1}{4}$$ - 1 = $$\frac{1}{4}$$

Solution:

 $$\eqalign{ & sec\theta = x + \frac{1}{{4x}} \cr & {\text{tan}}\theta = \sqrt {{\text{sec}}\theta - 1} \cr & {\text{tan}}\theta = \sqrt {{{\left[ {\frac{{4{x^2} + 1}}{{4x}}} \right]}^2} - 1} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} + 1} \right)}^2} - {{\left( {4x} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 - 8{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \frac{{\left( {4{x^2} - 1} \right)}}{{4x}} \cr & \therefore sec\theta + {\text{tan}}\theta \cr & = \frac{{4{x^2} + 1}}{{4x}} + \frac{{4{x^2} - 1}}{{4x}} \cr & = \frac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}} \cr & = \frac{{8{x^2}}}{{4x}} \cr & = 2x \cr & \cr & {\bf{Alternate:}} \cr & sec\theta = x + \frac{1}{{4x}} \cr & {\text{Put }}x = 1 \cr & sec\theta = 1 + \frac{1}{4} = \frac{5}{4} = \frac{{\text{H}}}{{\text{B}}} \cr & \tan \theta = \frac{{\text{P}}}{{\text{B}}} = \frac{3}{4} \cr & {\text{Now, }} \cr & sec\theta + {\text{tan}}\theta \cr & = \frac{5}{4} + \frac{3}{4} \cr & = \frac{{5 + 3}}{4} \cr & = \frac{8}{4} \cr & = 2 \times 1 \cr & = 2x\left( {x = 1} \right) \cr} $$

Solution:

 $$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$

Solution:

 $$\eqalign{ & {a^2}{\sec ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2}\left( {1 + ta{n^2}x} \right) - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + {a^2}{\tan ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + ta{n^2}x\left( {{a^2} - {b^2}} \right) = {c^2} \cr & \Rightarrow {a^2} - {c^2} = {\tan ^2}x\left( {{b^2} - {a^2}} \right) \cr & \Rightarrow \frac{{{a^2} - {c^2}}}{{{b^2} + {a^2}}} = {\text{ta}}{{\text{n}}^2}x \cr & \Rightarrow {\sec ^2}x - {\tan ^2}x = 1 \cr & \Rightarrow {\sec ^2}x = {\tan ^2}x + 1 \cr & \Rightarrow 1 + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {a^2} + {a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & {\sec ^2}x + {\tan ^2}x \cr & = \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & = \frac{{{b^2} + {a^2} - 2{c^2}}}{{{b^2} - {a^2}}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Arc length}} = \frac{\theta }{{180}} \times \pi r \cr & \Rightarrow 44 = \frac{{18}}{{180}} \times \frac{{22}}{7} \times r \cr & \Rightarrow r = 140\,{\text{m}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given, }}\frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = {\text{2}}\frac{{51}}{{79}} \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = \frac{{209}}{{79}} \cr & {\text{By componendo dividendo}} \cr & \left[ {\frac{a}{b} = \frac{c}{d},{\text{ }}\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}} \right] \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta + {\text{sec}}\theta - {\text{tan}}\theta }}{{{\text{sec}}\theta + {\text{tan}}\theta - {\text{sec}}\theta + {\text{tan}}\theta }} = \frac{{209 + 79}}{{209 - 79}} \cr & \Rightarrow \frac{{2sec\theta }}{{2{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{sec\theta }}{{{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{\frac{1}{{{\text{cos}}\theta }}}}{{\frac{{\sin \theta }}{{{\text{cos}}\theta }}}} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{1}{{\sin \theta }} = \frac{{288}}{{130}} \cr & \Rightarrow {\text{Therefore, }}\sin \theta = \frac{{130}}{{288}} \cr & \Rightarrow \sin \theta = \frac{{65}}{{144}} \cr} $$

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:

 $$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$