Practice MCQ Questions and Answer on Trigonometry

Solution:

 If in the any question componendo and dividendo already used as $$\frac{{a + b}}{{a - b}} = \frac{m}{n}$$ If second time you also want to apply componendo dividendo rule then result will be $$\eqalign{ & \frac{a}{b} = \frac{{m + n}}{{m - n}} \cr & \Leftrightarrow \frac{{\sin \theta + \cos \theta }}{{{\text{sin}}\theta - \cos \theta }} = 3 \cr & \Leftrightarrow \sin \theta + {\text{cos}}\theta = 3\sin \theta - 3{\text{cos}}\theta \cr & \Leftrightarrow 2\sin \theta = 4{\text{cos}}\theta \cr & \Leftrightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} = \frac{2}{1} \cr & \Leftrightarrow {\text{tan}}\theta = \frac{2}{1} = \frac{{\text{P}}}{{\text{B}}} \cr & \left[ {\therefore {\text{sin}}\theta = \frac{{\text{P}}}{{\text{H}}} = \frac{2}{{\sqrt 5 }}{\text{ and cos}}\theta = \frac{{\text{B}}}{{\text{H}}} = \frac{1}{{\sqrt 5 }}} \right] \cr} $$ $$\eqalign{ & \Leftrightarrow {\sin ^4}\theta - {\text{co}}{{\text{s}}^4}\theta \cr & \Leftrightarrow \left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right)\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow 1\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Leftrightarrow \frac{4}{5} - \frac{1}{5} \cr & \Leftrightarrow \frac{3}{5} \cr} $$

Solution:

 $$\eqalign{ & \tan \theta + \sin \theta = m \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta + 2{\text{ tan}}\theta .\sin \theta = {m^2}\,....(i) \cr & {\text{tan}}\theta - \sin \theta = n \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta - 2{\text{ tan}}\theta .\sin \theta = {n^2}\,....(ii) \cr & {\text{Substract from (i) and (ii)}} \cr & {m^2} - {n^2} = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta - {\text{ta}}{{\text{n}}^2}\theta - {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta \cr & {m^2} - {n^2} = 4{\text{tan}}\theta \sin \theta \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta {{\sin }^2}\theta } \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta - {{\sin }^2}\theta } \cr & = 4\sqrt {mn} \cr} $$

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:

 $$\eqalign{ & {\text{AC}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & {\text{sin A}} + \operatorname{cotC} \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} + \frac{{{\text{BC}}}}{{{\text{AB}}}} \cr & \frac{1}{{\sqrt 5 }} + \frac{1}{2} \cr & \Rightarrow \frac{{2 + \sqrt 5 }}{{2\sqrt 5 }} \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\tan }^6}\theta - {{\sec }^6}\theta + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{{{\left( {{{\tan }^2}\theta } \right)}^3} - {{\left( {{{\sec }^2}\theta } \right)}^3} + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right] \cr & = \frac{{\left( {{{\tan }^2}\theta - {{\sec }^2}\theta } \right)\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta + 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta - 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta - 2{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1}}{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)}^2}}} \cr & = - {\left( {\cos \theta \sin \theta } \right)^2} \cr & = - {\cos ^2}\theta {\sin ^2}\theta \cr} $$

Solution:

 cos245° + cos2135° + cos2225° + cos2315° = (cos245° + cos2315°) + (cos2135° + cos2225°)   [Here cos2α + cos2β = 1, where α + β = 360°] = 1 + 1 = 2

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:

 $$\eqalign{ & \left( {2\sin \theta + 3\cos \theta } \right) \cr & {\text{Maximum value of}} \cr & a\sin \theta + b\cos \theta \cr & = \sqrt {{a^2} + {b^2}} \cr & = \sqrt {{2^2} + {3^2}} \cr & = \sqrt {4 + 9} \cr & = \sqrt {13} \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\cos }^2}\theta }}{{{{\cot }^2}\theta - {{\cos }^2}\theta }} = 3 \cr & \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta \left( {\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^2}\theta }}} \right)}} = 3 \cr & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 3 \cr & \tan \theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \cot \theta + {\text{cosec}}\,\theta \cr & = \cot {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{{\sqrt 3 }} \cr & = \sqrt 3 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - A} \right) + \cot \left( {{{90}^ \circ } - A} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2A} \right)} \right]}} \cr & {\text{By putting }}A = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right) + \cot \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2 \times {{45}^ \circ }} \right)} \right]}} \cr & \Rightarrow \frac{{{{\left[ {\tan {{45}^ \circ } + \cot {{45}^ \circ }} \right]}^2}}}{{2{{\sec }^2}\left( {{{90}^ \circ } - {{90}^ \circ }} \right)}} \cr & \Rightarrow \frac{{{{\left[ {1 + 1} \right]}^2}}}{{2{{\sec }^2}{0^ \circ }}} \cr & \Rightarrow \frac{4}{{2 \times 1}} \cr & \Rightarrow 2 \cr} $$