Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & x\cos \theta - y\sin \theta = \sqrt {{x^2} + {y^2}} \,.....(i) \cr & \frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}}\,.....(ii) \cr & \frac{x}{{\sqrt {{x^2} + {y^2}} }}\cos \theta + \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }}\sin \theta = 1 \cr & {\text{from equation (i)}} \cr & \Rightarrow \sin \theta = \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }} \cr & \Rightarrow \cos \theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr & {\text{Put value in equation (ii)}} \cr & \therefore \frac{{{\text{co}}{{\text{s}}^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{\left( {{x^2} + {y^2}} \right){a^2}}} + \frac{{{y^2}}}{{\left( {{x^2} + {y^2}} \right){b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = \frac{1}{3}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1{\text{ }}\left( {{\text{Property}}} \right) \cr & \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = \frac{1}{3} \times 1 \cr & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{1}{3} \cr & \therefore \left( {\left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right) = {a^4} - {b^4}} \right) \cr} $$

$$\eqalign{ & \sin \theta + \cos \theta = \sqrt 2 \sin \left( {{{90}^ \circ } - \theta } \right) \cr & \sin \theta + \cos \theta = \sqrt 2 cos\theta \cr & {\text{Divide both sides by cos}}\theta \cr & {\text{tan}}\theta + 1 = \sqrt 2 \cr & \cot \theta = \frac{1}{{\sqrt 2 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 + 1 \cr} $$

$$\eqalign{ & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & {\text{Find }}{\sin ^4}\theta = ? \cr & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & \left( {{\text{by C & D}}} \right) \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{3 + 1}}{{3 - 1}} \cr & \Rightarrow {\text{tan}}\theta = 2 \cr & {\text{tan}}\theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{1} \cr & \Rightarrow {\sin ^4}\theta \Rightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$

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

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow \frac{1}{2} \times \cot {45^ \circ } \cr & \left[ {\frac{{1 + \left( {\sec {{45}^ \circ } + \tan {{45}^ \circ }} \right)}}{{{\text{cosec }}{{45}^ \circ }\left( {\sec {{45}^ \circ } - \tan {{45}^ \circ }} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + {{\left( {\sqrt 2 - 1} \right)}^2}}}{{\sqrt 2 \times \left( {\sqrt 2 - 1} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + 2 + 1 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{4 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2} \times 2\left[ {\frac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & = \frac{{\sin \theta .\operatorname{cosec} \theta .\tan \theta .\cot \theta }}{{{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta }} \cr & = \frac{{\sin \theta \times \frac{1}{{\sin \theta }} \times \tan \theta \times \frac{1}{{\tan \theta }}}}{1} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{cos}}\theta + \sec \theta = \sqrt 3 \cr & {\text{Cubing both sides}} \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3{\text{cos}}\theta \sec \theta \left( {{\text{cos}}\theta + \sec \theta } \right) = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3\sqrt 3 = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta = 0 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\sin ^2}{2^ \circ } + {\sin ^2}{4^ \circ } + {\sin ^2}{6^ \circ } + ..... + {\sin ^2}{90^ \circ } \cr & {\text{Number of terms}} \cr & = \frac{{l - a}}{d} + 1 \cr & = \frac{{90 - 2}}{2} + 1 \cr & = 45 \cr & {\text{But }}{\sin ^2}{90^ \circ } = 1 \cr & {\text{So, 22 pairs}} + {\sin ^2}{90^ \circ } \cr & = 22 + 1 \cr & = 23 \cr} $$

$$\eqalign{ & {\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right) = {\text{B}} \times \tan \left( {\theta - {{60}^ \circ }} \right) \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {\theta - {{60}^ \circ }} \right)}}{{\tan \left( {\theta + {{150}^ \circ }} \right)}} \cr & {\text{Put }}\theta = {90^ \circ } \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right)}}{{\tan \left( {{{90}^ \circ } + {{150}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan \left( {{{180}^ \circ } + {{60}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{1}{3} \cr & {\text{then, }}\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - \frac{4}{2} \cr & \Rightarrow \frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - 2 \cr & \Rightarrow \frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}} = - \frac{1}{2} \cr & {\text{Put in option (i)}} \cr & - \frac{{\sin {{90}^ \circ }}}{2} = - \frac{1}{2} \cr & {\text{So, option (A) is correct }} \cr} $$