Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{ }}\sqrt 3 \tan \theta = 3\sin \theta \cr & {\bf{Shortcut method:}} \cr & \Rightarrow {\text{ }}\sqrt 3 \frac{{\sin \theta }}{{\cos \theta }} = 3\sin \theta \cr & \Rightarrow \frac{{\sqrt 3 }}{{\cos \theta }} = 3 \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{3} \cr & {\text{then perpendicular}} = \sqrt 6 \cr} $$

$$\eqalign{ & \Rightarrow \left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow {\left( {\frac{P}{H}} \right)^2} - {\left( {\frac{B}{H}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{3}} \right)^2} \cr & \Rightarrow \frac{6}{9} - \frac{3}{9} \cr & \Rightarrow \frac{1}{3} \cr} $$

$$\sin \theta = \frac{a}{b} = \frac{{\text{P}}}{{\text{H}}}$$

$$\eqalign{ & {\text{BC}} = \sqrt {{b^2} - {a^2}} \cr & \left[ {{\text{using pythagorad theorem}}} \right] \cr & \therefore sec\theta - \cos \theta \cr & = \frac{{\text{H}}}{{\text{B}}} - \frac{{\text{B}}}{{\text{H}}} \cr & = \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & = \frac{b}{{\sqrt {{b^2} - {a^2}} }} - \frac{{\sqrt {{b^2} - {a^2}} }}{b} \cr & = \frac{{{b^2} - {{\left( {\sqrt {{b^2} - {a^2}} } \right)}^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr & = \frac{{{b^2} - {b^2} + {a^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr & = \frac{{{a^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr} $$

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

$$\eqalign{ & sec\theta + {\text{tan}}\theta = 2 + \sqrt 5 \,.....(i) \cr & {\text{se}}{{\text{c}}^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \left( {sec\theta - {\text{tan}}\theta } \right)\left( {sec\theta + {\text{tan}}\theta } \right) = 1 \cr & \left( {sec\theta - {\text{tan}}\theta } \right) = \frac{1}{{2 + \sqrt 5 }} = \frac{1}{{\sqrt 5 + 2}} = \sqrt 5 - 2\,\,.....(ii) \cr & {\text{Adding equation }}{\text{ (i) and (ii)}} \cr & {\text{2}}sec\theta = 2 + \sqrt 5 + \sqrt 5 - 2 \cr & \Rightarrow 2sec\theta = 2\sqrt 5 \cr & \Rightarrow sec\theta = \sqrt 5 \cr & \Rightarrow {\text{cos}}\theta = \frac{1}{{\sqrt 5 }} \cr & \Rightarrow {\sin ^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = 1 - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Rightarrow {\sin ^2}\theta = \frac{4}{5} \cr & \Rightarrow \sin \theta = \frac{2}{{\sqrt 5 }} \cr & \therefore \sin \theta + {\text{cos}}\theta = \frac{2}{{\sqrt 5 }} + \frac{1}{{\sqrt 5 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 5 }} \cr} $$

$$\eqalign{ & sec\alpha + \tan \alpha = 2\,.....(i) \cr & sec\alpha - \tan \alpha = \frac{1}{2}\,.....(ii) \cr & {\text{Adding equation (i) and (ii)}} \cr & {\text{2sec}}\alpha = 2 + \frac{1}{2} \cr & \sec \alpha = \frac{{5 \to {\text{H}}}}{{4 \to {\text{B}}}} \cr} $$

$$\sin \alpha = \frac{3}{5} = 0.6$$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{cosec A}} + {\text{cot A}} = 3 \cr & {\text{cosec A}} - {\text{cot A}} = \frac{1}{3} \cr & \overline {2{\text{cosec A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{3}} \cr & {\text{cosec A}} = \frac{{10}}{6} \cr & {\text{sin A}} = \frac{6}{{10}} \cr & {\text{sin A}} = \frac{3}{5} \cr} $$

$$\eqalign{ & {\text{3}}\left( {{{\sec }^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right) = 5 \cr & {\sec ^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = \frac{5}{3}\,.....(i) \cr & {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1\,......(ii) \cr & {\text{Adding equation (i) and (ii)}} \cr & {\text{2}}{\sec ^2}\theta = \frac{8}{3} \cr & sec\theta = \frac{2}{{\sqrt 3 }} \cr & \therefore \theta = {30^ \circ } \cr & \cos 2\theta = \cos 2\left( {{{30}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos {60^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$

$$\eqalign{ & \angle {\text{A}} + \angle {\text{B}} = {135^ \circ }\,.....(i) \cr & \angle {\text{A}} - \angle {\text{B}} = \frac{\pi }{{12}} = {15^ \circ }\,.....(ii) \cr & Adding{\text{ both equations}} \cr & \Rightarrow 2\angle {\text{A}} = {150^ \circ } \cr & \Rightarrow \angle {\text{A}} = {75^ \circ } \cr & \therefore \angle {\text{B}} = {60^ \circ } \cr} $$

$$\eqalign{ & {\sin ^{12}} + 3{\sin ^{10}}x + 3{\sin ^8}x + {\sin ^6}x - 1 \cr & \Rightarrow {\left( {{{\sin }^4}x + {{\sin }^2}x} \right)^3} - 1 \cr & \Rightarrow {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)^3} - 1 \cr & \left[ {\cos x + {{\cos }^2}x = 1} \right] \cr & (\cos x = 1 - {\cos ^2}x = {\sin ^2}x) \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {\text{cos}}\theta + \sin \theta = \sqrt 2 \cos \theta \cr & {\text{Squaring both sides}} \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta + 2\cos \theta .\sin \theta = 2{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2{\text{cos}}\theta .{\text{sin}}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\sqrt 2 \cos \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \cos \theta - \sin \theta = \frac{{2\sin \theta .\cos \theta }}{{\sqrt 2 \cos \theta }} \cr & \Rightarrow \sqrt 2 \sin \theta \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}\sqrt 2 \cos \theta = \alpha \cr & \therefore \cos \theta \pm \sin \theta = a \cr & \cos \theta \pm \sin \theta = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - 2{\text{co}}{{\text{s}}^2}\theta } \cr & = \sqrt {2\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = \sqrt {2{{\sin }^2}\theta } \cr & = \sqrt 2 \sin \theta \cr} $$