Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & b = \frac{c}{{1 + \sin x}} \cr & {\text{Go through option from option B}} \cr & b = \frac{{1 + \sin x - \cos x}}{{1 + \sin x}} \cr & = \frac{{{{\left( {1 + \sin x} \right)}^2} - {{\cos }^2}x}}{{1 + \sin x\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{1 + {{\sin }^2}x + 2{{\sin }^2}x - 1 + {{\sin }^2}x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2{{\sin }^2}x + 2\sin x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2\sin x\left( {\sin x + 1} \right)}}{{1 + \sin x + \cos x}} \cr & = \frac{{2\sin x}}{{1 + \sin x + \cos x}} \cr & = a \cr} $$

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

A + B = C
tan(A + B) = tanC
$$\frac{{\tan {\text{A}} + \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} = \tan {\text{C}}$$
tanA + tanB = tanC - tanAtanBtanC
tanAtanBtanC = tanC - tanA - tanB

cos²45° + cos²135° + cos²225° + cos²315°
= (cos²45° + cos²315°) + (cos²135° + cos²225°)   [Here cos²α + cos²β = 1, where α + β = 360°]
= 1 + 1
= 2

$$\eqalign{ & 3{\sin ^2}{30^ \circ } + \frac{3}{5}{\cos ^2}{60^ \circ } - 2{\sec ^2}{45^ \circ } \cr & = 3 \times \frac{1}{4} + \frac{3}{5} \times \frac{1}{4} - 2 \times 2 \cr & = \frac{3}{4} + \frac{1}{5} \times \frac{3}{4} - 4 \cr & = \frac{3}{4} + \frac{3}{{20}} - 4 \cr & = \frac{{15 + 3 - 80}}{{20}} \cr & = \frac{{ - 62}}{{10}} \cr & = \frac{{ - 31}}{{10}} \cr} $$

$$\eqalign{ & 1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}} \cr & = 1 + \frac{{{{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{1 + \sec A + {{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{\sec A\left( {1 + \sec A} \right)}}{{1 + \sec A}} \cr & = \sec A \cr} $$

(sinA + cosA)(1 - sinA.cosA)
= (sinA + cosA)[sin²A + cos²A - sinA.cosA]
= (sin³A + cos³A)

$$\eqalign{ & \sin \left( {180 - \theta } \right)\sin \left( {90 - \theta } \right) + \frac{{\cot \left( {90 - \theta } \right)}}{{1 + {{\tan }^2}\theta }} \cr & \Rightarrow \sin \theta \cos \theta + \frac{{\tan \theta }}{{{{\sec }^2}\theta }} \cr & \Rightarrow \sin \theta \cos \theta + \frac{{\sin \theta }}{{\cos \theta }} \times {\cos ^2}\theta \cr & \Rightarrow \sin \theta \cos \theta + \sin \theta \cos \theta \cr & \Rightarrow 2\cos \theta \sin \theta \cr} $$

$$\eqalign{ & \frac{{\left( {\sin 4x + \sin 4y} \right)\left[ {\tan \left( {2x - 2y} \right)} \right]}}{{\sin 4x - \sin 4y}} \cr & \Rightarrow \frac{{2\sin \left( {\frac{{4x + 4y}}{2}} \right)\cos \left( {\frac{{4x - 4y}}{2}} \right)\left[ {\frac{{\sin \left( {2x - 2y} \right)}}{{\cos \left( {2x - 2y} \right)}}} \right]}}{{2\cos \left( {\frac{{4x + 4y}}{2}} \right)\sin \left( {\frac{{4x - 4y}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {2x + 2y} \right)\cos \left( {2x - 2y} \right)}}{{\cos \left( {2x + 2y} \right)\sin \left( {2x - 2y} \right)}} \times \left[ {\frac{{\sin \left( {2x - 2y} \right)}}{{\cos \left( {2x - 2y} \right)}}} \right] \cr & \Rightarrow \tan \left( {2x + 2y} \right) \cr} $$

cos⁴α - sin⁴α = $$\frac{5}{6}$$
(cos²α + sin²α)(cos²α - sin²α) = $$\frac{5}{6}$$
1 × (cos²α - 1 + cos²α) = $$\frac{5}{6}$$
2cos²α - 1 = $$\frac{5}{6}$$