Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \frac{{3\left( {{{\cot }^2}{{47}^ \circ } - {{\sec }^2}{{43}^ \circ }} \right) - 2\left( {{{\tan }^2}{{23}^ \circ } - {\text{cose}}{{\text{c}}^2}{{67}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}\left( {{{68}^ \circ } + \theta } \right) - \tan \left( {\theta + {{61}^ \circ }} \right) - {{\tan }^2}\left( {{{22}^ \circ } - \theta } \right) + \cot \left( {{{29}^ \circ } - \theta } \right)}} \cr & = \frac{{3\left( {{{\tan }^2}{{43}^ \circ } - {{\sec }^2}{{43}^ \circ }} \right) - 2\left( {{{\tan }^2}{{23}^ \circ } - {{\sec }^2}{{23}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}\left( {{{68}^ \circ } + \theta } \right) - \tan \left( {\theta + {{61}^ \circ }} \right) - {{\cot }^2}\left( {{{68}^ \circ } - \theta } \right) + \tan \left( {{{61}^ \circ } - \theta } \right)}} \cr & = \frac{{3 \times \left( { - 1} \right) - 2 \times \left( { - 1} \right)}}{1} \cr & = \frac{{ - 3 + 2}}{1} \cr & = - 1 \cr} $$

In these type of questions we can go through option-
From option B
$$\eqalign{ & \frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right) + \cos A}} = 1 - \cos A + \sin A \cr & 2\sin A\left( {1 + \sin A} \right) = {\left( {1 + \sin A} \right)^2} - {\cos ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 1 + \sin 2A + 2\sin A - 1 + {\sin ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 2\sin A\left( {1 + \sin A} \right) \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put }}A = {90^ \circ } \cr & \frac{{\left( {2\sin {{90}^ \circ }} \right)\left( {1 + \sin {{90}^ \circ }} \right)}}{{1 + \sin {{90}^ \circ } + \cos {{90}^ \circ }}} = 2 \cr & {\text{Now from option B}} \cr & 1 + \sin A - \cos A \cr & = 1 + \sin {90^ \circ } - \cos {90^ \circ } \cr & = 1 + 1 - 0 \cr & = 2 \cr} $$

$$\eqalign{ & \frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7} \cr & 7\sec \theta - 7\tan \theta = \sec \theta + \tan \theta \cr & 6\sec \theta = 8\tan \theta \cr & \sin \theta = \frac{6}{8} = \frac{3}{4} = \frac{P}{H} \cr & B = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & \Rightarrow \frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }} \cr & = \frac{{\frac{H}{P} + {{\left( {\frac{B}{P}} \right)}^2}}}{{\frac{H}{P} - {{\left( {\frac{B}{P}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}}{{\frac{4}{3} - {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + \frac{7}{9}}}{{\frac{4}{3} - \frac{7}{9}}} \cr & = \frac{{\frac{{12 + 7}}{9}}}{{\frac{{12 - 7}}{9}}} \cr & = \frac{{19}}{5} \cr} $$

$$\eqalign{ & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \cos \left( {{{12}^ \circ } - \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}{{20}^ \circ }} \right)}}{{\sin {{25}^ \circ }\cos {{65}^ \circ } + \cos {{25}^ \circ }\sin {{65}^ \circ }}} \cr & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \sin \left( {{{78}^ \circ } + \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {{\sec }^2}{{70}^ \circ }} \right)}}{{\sin {{25}^ \circ }\sin {{25}^ \circ } + \cos {{25}^ \circ }\sin {{25}^ \circ }}} \cr & \frac{{ - 1}}{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}} \cr & = - 1 \cr} $$

$$\eqalign{ & \frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{2 \times \sqrt 3 }}{{1 + 3}} \cr & = \frac{{\sqrt 3 }}{2} \cr & = \sin {60^ \circ } \cr} $$

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

$$\eqalign{ & \frac{{4{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }{{\cos }^2}{{45}^ \circ } + {{\sec }^2}{{48}^ \circ } - {{\cot }^2}{{42}^ \circ }}}{{\cos {{37}^ \circ }\sin {{53}^ \circ } + \sin {{37}^ \circ }\cos {{53}^ \circ } + \tan {{18}^ \circ }\tan {{72}^ \circ }}} \cr & = \frac{{4{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {\text{cose}}{{\text{c}}^2}{{42}^ \circ } - {{\cot }^2}{{42}^ \circ }}}{{\cos {{37}^ \circ }\cos {{37}^ \circ } + \sin {{37}^ \circ }\sin {{37}^ \circ } + 1}} \cr & = \frac{{4 \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{2} + 1}}{{{{\cos }^2}{{37}^ \circ } + {{\sin }^2}{{37}^ \circ } + 1}} \cr & = \frac{{\frac{4}{3} + \frac{1}{8} + 1}}{{1 + 1}} \cr & = \frac{{32 + 3 + 24}}{{24 \times 2}} \cr & = \frac{{59}}{{48}} \cr} $$

4sin²θ = 3(1 + cosθ)
4(1 - cos²θ) = 3 + 3cosθ
4 - 4cos²θ = 3 + 3cosθ
4cos²θ + 3cosθ - 1 = 0
4cos²θ + 4cosθ - cosθ - 1 = 0
4cosθ(cosθ + 1) - 1(cosθ + 1) = 0
(4cosθ -1)(cosθ + 1) = 0
4cosθ - 1 = 0
cosθ = $$\frac{1}{4}$$

Then, 2tanθ + 4tanθ - secθ
$$\eqalign{ & = 2 \times \frac{{\sqrt {15} }}{1} + 4 \times \frac{{\sqrt {15} }}{4} - \frac{4}{1} \cr & = 2\sqrt {15} + \sqrt {15} - 4 \cr & = 3\sqrt {15} - 4 \cr} $$

(cos⁶θ + sin⁶θ - 1)(tan²θ + cot²θ + 2) + 3
Put θ = 45°
= (cos⁶45° + sin⁶45° - 1)(tan²45° + cot²45° + 2) + 3
$$\eqalign{ & = \left( {\frac{1}{8} + \frac{1}{8} - 1} \right)\left( {1 + 1 + 2} \right) + 3 \cr & = - \frac{3}{4} \times 4 + 3 \cr & = 0 \cr} $$

$$\eqalign{ & 5\sin \theta - 4\cos \theta = 0 \cr & 5\sin \theta = 4\cos \theta \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{5} \cr & \tan \theta = \frac{4}{5} \cr & \frac{{5\sin \theta - 2\cos \theta }}{{5\sin \theta + 3\cos \theta }} \cr & = \frac{{5\tan \theta - 2}}{{5\tan \theta + 3}} \cr & = \frac{{5 \times \frac{4}{5} - 2}}{{5 \times \frac{4}{5} + 2}} \cr & = \frac{2}{7} \cr} $$