Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \Rightarrow x\tan {60^ \circ } + \cos {45^ \circ } = \sec {45^ \circ } \cr & \Rightarrow x.\sqrt 3 + \frac{1}{{\sqrt 2 }} = \sqrt 2 \cr & \Rightarrow x\sqrt 6 + 1 = 2 \cr & \Rightarrow x = \frac{1}{{\sqrt 6 }} \cr }$$
Squaring both sides and addition 1 both sides
$$\eqalign{ & \Rightarrow {x^2} + 1 = {\left( {\frac{1}{{\sqrt 6 }}} \right)^2} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{1}{6} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{7}{6} \cr} $$

sin(2x - 20) = cos(2y + 20)
⇒ sin(2x - 20) = sin(90 - (2y + 20))
i.e. 2x - 20 = 90 - 2y - 20
⇒ 2x + 2y = 90
⇒ x + y = 45°
∴ tan(x + y) = tan 45° = 1

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

$$\eqalign{ & {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta = 1\,......({\text{i}}) \cr & \because {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {{{\sec }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{{{{\sec }^2}\theta }} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{co}}{{\text{s}}^2}\theta \cr & \because {\text{ co}}{{\text{s}}^4}\theta + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{co}}{{\text{s}}^2}\theta } \right)^2} + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{ta}}{{\text{n}}^2}\theta } \right)^2} + {\text{ta}}{{\text{n}}^2}\theta \cr & = {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta \cr & = 1{\text{ from equation }}\left( {\text{i}} \right) \cr} $$

$$\eqalign{ & {\text{8}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{12}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) \cr & {\text{Put }}\theta = {0^ \circ } \cr & = 8\left( {{{\sin }^6}{0^ \circ } + {{\cos }^6}{0^ \circ }} \right) - 12\left( {{{\sin }^4}{0^ \circ } + {{\cos }^4}{0^ \circ }} \right) \cr & = 8\left( {0 + 1} \right) - 12\left( {0 + 1} \right) \cr & = - 4 \cr} $$

$$\eqalign{ & \sec A = \frac{{17 \to H}}{{8 \to B}},\,A < {90^ \circ } \cr & {L^2} = {H^2} - {B^2} \cr & {L^2} = 289 - 64 \cr & {L^2} = 225 \cr & L = 15 \cr & {\text{Then}}, \cr & \frac{{34\sin A + 15\cot A}}{{68\cos A - 16\tan A}} \cr & = \frac{{34 \times \frac{{15}}{{17}} + 15 \times \frac{8}{{15}}}}{{68 \times \frac{8}{{17}} - 16 \times \frac{{15}}{8}}} \cr & = \frac{{30 + 8}}{{32 - 30}} \cr & = \frac{{38}}{2} \cr & = 19 \cr} $$

$$\eqalign{ & \frac{{\cos \left( {\pi - A} \right).\cot \left( {\frac{\pi }{2} + A} \right)\cos \left( { - A} \right)}}{{\tan \left( {\pi + A} \right)\tan \left( {\frac{{3\pi }}{2} + A} \right)\sin \left( {2\pi - A} \right)}}\, \cr & = \frac{{\left( { - \cos A} \right) \times \left( { - \tan A} \right) \times \cos A}}{{\tan A \times \left( { - \cot A} \right) \times \left( { - \sin A} \right)}} \cr & = \frac{{{{\cos }^2}A}}{{\frac{{\cos A}}{{\sin A}} \times \sin A}} \cr & = \frac{{{{\cos }^2}A}}{{\cos A}} \cr & = \cos A \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\bf{Formula \,used:}} \cr & {\text{Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & {\text{Where, h = Hypotenuse, p = Perpendicular and b = base}} \cr & {\bf{Calculation:}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\text{Let, }}\tan A = x \cr & \Rightarrow {x^2} + 2x - 63 = 0 \cr & \Rightarrow {x^2} + 9x - 7x - 63 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 7\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 7} \right) \cr & \Rightarrow x + 9 = 0 \Rightarrow x = - 9\left[ {'' - ''{\text{ will be neglected because }}0 < A < \frac{\pi }{2}} \right] \cr & \Rightarrow x - 7 = 0 \Rightarrow x = 7 \cr & {\text{So,}}\tan A = \frac{7}{1} = \frac{p}{b} \cr & {\text{Using Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & h = \sqrt {{7^2} + {1^2}} \cr & h = \sqrt {50} \cr & {\text{So}},\,2\sin A + 5\cos A \cr & = \left[ {2 \times \frac{7}{{\sqrt {50} }}} \right] + \left[ {5 \times \frac{1}{{\sqrt {50} }}} \right] \cr & = \frac{{14}}{{\sqrt {50} }} + \frac{5}{{\sqrt {50} }} \cr & = \frac{{14 + 5}}{{\sqrt {50} }} \cr & = \frac{{19}}{{\sqrt {50} }} \cr & \therefore {\text{The value of}}\left( {2\sin A + 5\cos A} \right){\text{is }}\frac{{19}}{{\sqrt {50} }}. \cr} $$

$$\eqalign{ & \frac{{{{\sin }^2}\phi - 3\sin \phi + 2}}{{{{\cos }^2}\phi }} = 1 \cr & \Rightarrow {\sin ^2}\phi - 3\sin \phi + 2 = {\cos ^2}\phi \cr & \Rightarrow {\sin ^2}\phi - 3\sin \phi + 2 = 1 - {\sin ^2}\phi \cr & \Rightarrow 2{\sin ^2}\phi - 3\sin \phi + 1 = 0 \cr & \Rightarrow 2{\sin ^2}\phi - 2\sin \phi - \sin \phi + 1 = 0 \cr & \Rightarrow 2\sin \phi \left( {\sin \phi - 1} \right) - 1\left( {\sin \phi - 1} \right) = 0 \cr & \Rightarrow \left( {\sin \phi - 1} \right)\left( {2\sin \phi - 1} \right) = 0 \cr & \therefore \sin \phi = 1\,\,\left( {{\text{not valid}}} \right) \cr & {\text{and }}\sin \phi = \frac{1}{2} \cr & \Rightarrow \sin \phi = \sin {30^ \circ } \cr & \therefore \phi = {30^ \circ } \cr & \therefore \cos 2\phi - \sin 3\phi + {\text{cosec }}2\phi \cr & = \cos {60^ \circ } - \sin {90^ \circ } + {\text{cosec }}{60^ \circ } \cr & = \frac{1}{2} - 1 + \frac{2}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} - \frac{1}{2} \cr & = \frac{{\left( {4 - \sqrt 3 } \right)}}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{4\sqrt 3 - 3}}{6} \cr & \cr & {\bf{Alternative:}} \cr & {\text{Put }}\phi = {30^ \circ } \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \frac{{{{\sin }^2}{{30}^ \circ } - 3\sin {{30}^ \circ } + 2}}{{{{\cos }^2}{{30}^ \circ }}} \cr & = \frac{{\frac{1}{4} - \frac{3}{2} + 2}}{{\frac{3}{4}}} \cr & = \frac{{1 - 6 + 8}}{3} \cr & = \frac{3}{3} \cr & = 1 = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \therefore \cos 2\phi - \sin 3\phi + {\text{cosec }}2\phi \cr & = \cos {60^ \circ } - \sin {90^ \circ } + {\text{cosec }}{60^ \circ } \cr & = \frac{1}{2} - 1 + \frac{2}{{\sqrt 3 }} \cr & = - \frac{1}{2} + \frac{2}{{\sqrt 3 }} \cr & = \frac{{\left( { - \sqrt 3 + 4} \right)}}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{ - 3 + 4\sqrt 3 }}{6} \cr} $$

$$\eqalign{ & \sin \theta = \sqrt 3 \cos \theta \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \theta = {60^ \circ } \cr & 2{\sin ^2}{60^ \circ } + {\sec ^2}{60^ \circ } + \tan {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 2 \times \frac{3}{4} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{2} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{{11}}{2} + \frac{5}{{\sqrt 3 }} \cr & = \frac{{11\sqrt 3 + 10}}{{2\sqrt 3 }} \cr & = \frac{{33 + 10\sqrt 3 }}{6} \cr} $$