Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

∵ P + Q + R = 60°
By putting P = 0°, Q = 0° and R = 60°
⇒ cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP)
⇒ 1 × cos60°(cos0° - sin0°) + sin0°.sin60°(sin0° - cos0°)
⇒ $$\frac{1}{2}$$ (1 - 0) + 0
⇒ $$\frac{1}{2}$$

$$\eqalign{ & \left[ {\frac{{{{\sin }^2}{{24}^ \circ } + {{\sin }^2}{{66}^ \circ }}}{{{{\cos }^2}{{24}^ \circ } + {{\cos }^2}{{66}^ \circ }}} + {{\sin }^2}{{61}^ \circ } + \cos {{61}^ \circ }\sin {{29}^ \circ }} \right] \cr & = \frac{1}{1} + {\sin ^2}{61^ \circ } + \cos {61^ \circ }\sin \left( {{{90}^ \circ } - {{61}^ \circ }} \right) \cr & = 1 + {\sin ^2}{61^ \circ } + {\cos ^2}{61^ \circ } \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\sin \alpha - \sin \beta = 2\cos \frac{{\alpha + \beta }}{2}.\sin \frac{{\alpha - \beta }}{2}$$

6tanA(tanA + 1) = 5 - tanA
6tan²A + 6tanA + tanA = 5
6tan²A + 7tanA = 5
6tan²A + 7tanA - 5 = 0
6tan²A + 10tanA - 3tanA - 5 = 0
2tanA(3tanA + 5) - 1(3tanA + 5) = 0
(2tanA - 1)(3tanA + 5) = 0
tanA = $$\frac{1}{2}$$ and tanA = $$ - \frac{5}{2}$$
taking tanA $$ = \frac{1}{2} = \frac{P}{B}$$
H = √5
$$\eqalign{ & \therefore \sin A + \cos A \cr & = \frac{P}{H} + \frac{B}{H} \cr & = \frac{1}{{\sqrt 5 }} + \frac{2}{{\sqrt 5 }} \cr & = \frac{3}{{\sqrt 5 }} \cr} $$

tan²A - 6tanA + 9 = 0
tan²A - 3tanA - 3tanA + 9 = 0
tanA(tanA - 3) - 3(tanA - 3) = 0
(tanA - 3)(tanA - 3) = 0
tanA $$ = \frac{3}{1} = \frac{P}{B}$$
H² = P² + B²
$$\eqalign{ & H = \sqrt {9 + 1} = \sqrt {10} \cr & 6\cot A + 8\sqrt {10} \cos A \cr & = 6\left( {\frac{B}{P}} \right) + 8\sqrt {10} \left( {\frac{B}{H}} \right) \cr & = 6\left( {\frac{1}{3}} \right) + 8\sqrt {10} \left( {\frac{1}{{\sqrt {10} }}} \right) \cr & = 2 + 8 \cr & = 10 \cr} $$

$$\eqalign{ & \sqrt 3 \tan \theta = 3\sin \theta \cr & \Rightarrow \frac{{\tan \theta }}{{\sin \theta }} = \frac{3}{{\sqrt 3 }} \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} \times \frac{1}{{\sin \theta }} = \frac{3}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\cos \theta }} = \sqrt 3 \cr & \Rightarrow \cos \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow {\cos ^2}\theta = \frac{1}{3} \cr & \therefore \,{\sin ^2}\theta - {\cos ^2}\theta \cr & = 1 - {\cos ^2}\theta - {\cos ^2}\theta \cr & = 1 - 2{\cos ^2}\theta \cr & = 1 - 2 \times \frac{1}{3} \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & \tan \theta + \cot \theta = - 2 \cr & \Rightarrow \tan \theta + \frac{1}{{\tan \theta }} = - 2 \cr & \Rightarrow \tan \theta = - 1 \cr & \Rightarrow {\tan ^9}\theta + {\cot ^9}\theta \cr & = {\tan ^9}\theta + \frac{1}{{{{\tan }^9}\theta }} \cr & = {\left( { - 1} \right)^9} + \frac{1}{{{{\left( { - 1} \right)}^9}}} \cr & = - 1 - 1 \cr & = - 2 \cr} $$

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

$$\eqalign{ & \frac{{3\sin {{58}^ \circ }}}{{\cos {{32}^ \circ }}} + \frac{{3\sin {{42}^ \circ }}}{{\cos {{48}^ \circ }}} \cr & = \frac{{3\cos {{32}^ \circ }}}{{\cos {{32}^ \circ }}} + \frac{{3\cos {{48}^ \circ }}}{{\cos {{48}^ \circ }}} \cr & \left\{ {\therefore \,{{58}^ \circ } + {{32}^ \circ } = {{90}^ \circ }\,\& \,{{42}^ \circ } + {{48}^ \circ } = {{90}^ \circ }} \right\} \cr & = 3 + 3 \cr & = 6 \cr} $$

(sin²θ + cos²θ)² = 1
sin⁴θ + cos⁴θ + 2sin²θ.cos²θ = 1
sin⁴θ + cos⁴θ = 1 - 2sin²θ.cos²θ
sin⁴θ + cos⁴θ = 1 - 2sin²θ(1 - sin²θ)
sin⁴θ + cos⁴θ = 1 - 2sin²θ + 2sin⁴θ