Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \cot \theta = 4 \cr & \therefore \frac{{5\sin \theta + 3\cos \theta }}{{5\sin \theta - 3\cos \theta }} \cr & = \frac{{5 + 3\cot \theta }}{{5 - 3\cot \theta }} \cr & = \frac{{5 + 3 \times 4}}{{5 - 3 \times 4}} \cr & = - \frac{{17}}{7} \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\sin ^2}{70^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = 1 \cr} $$

$$\eqalign{ & r\sin \theta = 1,{\text{ }}r\cos \theta = \sqrt 3 \cr & {\text{Put }}\theta = {30^ \circ } \cr & r = 2 \cr & {\text{So}},{\text{ }}{r^2}\tan \theta \cr & = {\left( 2 \right)^2} \times {\text{tan}}{30^ \circ } \cr & = 4 \times \frac{1}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \sec \theta + \tan \theta = m\,.......(i) \cr & {\text{then, }}\sec \theta - \tan \theta = \frac{1}{m}\,......(ii) \cr & Because{\text{ }}{\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & {\text{From equation (i)}} - {\text{(ii)}} \cr & 2\tan \theta = m - \frac{1}{m} \cr} $$

$$\eqalign{ & {\text{tan}}\theta = \frac{{{m^2} - 1}}{{2m}} \cr & \sin \theta = \frac{{{m^2} - 1}}{{{m^2} + 1}} \cr} $$

$$\eqalign{ & \frac{{\cot \theta + \operatorname{cosec} \theta - 1}}{{\cot \theta + \operatorname{cosec} \theta + 1}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \frac{{1 + \sqrt 2 - 1}}{{1 + \sqrt 2 + 1}} \cr & = \frac{{\sqrt 2 }}{{2 + \sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} \cr & = \frac{1}{{\sqrt 2 + 1}} \cr & = \sqrt 2 - 1 \cr & {\text{Now option B}} \cr & \frac{{1 - \cos \theta }}{{\sin \theta }} \cr & = \frac{{1 - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}} \cr & = \sqrt 2 - 1{\text{ }}\left( {{\text{Satisfy}}} \right) \cr} $$

$$\eqalign{ & {\text{secA}} + {\text{tanA}} = a \cr & {\text{we know that}} \cr & \Rightarrow {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1 \cr & \Rightarrow \left( {{\text{secA}} - {\text{tanA}}} \right)\left( {{\text{secA}} + {\text{tanA}}} \right) = 1 \cr & \Rightarrow {\text{secA}} - {\text{tanA}} = \frac{1}{a} \cr & \Rightarrow {\text{secA}} + {\text{tanA}} = a \cr & \Rightarrow {\text{2secA}} = a + \frac{1}{a} \cr & \Rightarrow {\text{2secA}} = \frac{{{a^2} + 1}}{a} \cr & \Rightarrow \sec \theta = \frac{{{a^2} + 1}}{{2a}} \cr & {\text{So, }}\cos \theta = \frac{{2a}}{{{a^2} + 1}} \cr} $$

$$\eqalign{ & {\text{sin P}} + {\text{cosec P}} = 2 \cr & {\text{For P}} = {90^ \circ } \cr & \Rightarrow {\text{sin }}{90^ \circ } + {\text{cosec }}{90^ \circ } = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & \Rightarrow 2 = 2\left( {{\text{satisfy}}} \right) \cr & {\text{So, }}{\sin ^7}{\text{P}} + {\text{cose}}{{\text{c}}^7}{\text{P}} \cr & \Rightarrow {\sin ^7}{90^ \circ } + {\text{cose}}{{\text{c}}^7}{90^ \circ } \cr & \Rightarrow {1^7} + {1^7} \cr & \Rightarrow 2 \cr} $$

$${\text{ 2}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right){\text{ + 1}}$$
$$ \Rightarrow {\text{2}}\left( {1 - 3{{\sin }^2}\theta {\text{co}}{{\text{s}}^2}\theta } \right) - $$     $${\text{3}}\left( {1 - 2{{\sin }^2}\theta c{\text{o}}{{\text{s}}^2}\theta } \right){\text{ + }}$$     $${\text{1}}$$
$$\eqalign{ & \Rightarrow {\text{ 2}} - 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta - 3 + 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta {\text{ + 1}} \cr & \Rightarrow 2 - 3 + 1 \cr & \Rightarrow 0 \cr} $$

$${\text{cos}}\theta = \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$

AC² = (x² + y²)² - (x² - y²)²
        = x⁴ + y⁴ + 2x²y² - x⁴ - y⁴ + 2x²y²
        = 4x²y²
⇒ AC = 2xy
⇒ cotθ = $$\frac{{{x^2} - {y^2}}}{{2xy}}$$

$$\eqalign{ & x = {\text{cosec}}\theta - \sin \theta {\text{ }} \cr & y = \sec \theta - \cos \theta \cr & {\text{Put }}\theta = {45^ \circ } \cr & x = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & y = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & {\text{by options (B) }}{x^2}{y^2}\left( {{x^2} + {y^2} + 3} \right) \cr & = {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \left[ {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 3} \right] \cr & = \frac{1}{2} \times \frac{1}{2}\left( {\frac{1}{2} + \frac{1}{2} + 3} \right) \cr & = \frac{1}{4}\left( {1 + 3} \right) \cr & = 1{\text{ }}\left( {{\text{satisfy}}} \right) \cr} $$