Trigonometrical Ratios and Identities 1 Question 2
2. The value of $\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}$ is
(a) $\frac{3}{2}\left(1+\cos 20^{\circ}\right)$
(b) $\frac{3}{4}+\cos 20^{\circ}$
(c) $3 / 2$
(d) $3 / 4$
(2019 Main, 9 April I)
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Answer:
Correct Answer: 2. (a)
Solution:
- We have, $\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}$
$$ \begin{aligned} & =\frac{1}{2}\left[2 \cos ^{2} 10^{\circ}-2 \cos 10^{\circ} \cos 50^{\circ}+2 \cos ^{2} 50^{\circ}\right] \\ & =\frac{1}{2}\left[1+\cos 20^{\circ}-\left(\cos 60^{\circ}+\cos 40^{\circ}\right)+1+\cos 100^{\circ}\right] \\ & {\left[\because 2 \cos ^{2} A=1+\cos 2 A\right. \text { and }} \\ & 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] \end{aligned} $$
$$ \begin{aligned} & =\frac{1}{2} 2+\cos 20^{\circ}+\cos 100^{\circ}-\frac{1}{2}-\cos 40^{\circ} \quad \because \cos 60^{\circ}=\frac{1}{2} \\ & =\frac{1}{2} \frac{3}{2}+\left(\cos 20^{\circ}-\cos 40^{\circ}\right)+\cos 100^{\circ} \\ & =\frac{1}{2} \frac{3}{2}-2 \sin \frac{20^{\circ}+40^{\circ}}{2} \sin \frac{20^{\circ}-40^{\circ}}{2}+\cos 100^{\circ} \\ & \quad \because \cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2} \\ & =\frac{1}{2} \frac{3}{2}-2 \sin 30^{\circ} \sin \left(-10^{\circ}\right)+\cos \left(90^{\circ}+10^{\circ}\right) \\ & =\frac{1}{2} \frac{3}{2}+\sin 10^{\circ}-\sin 10^{\circ}\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right] \\ & =\frac{1}{2} \times \frac{3}{2}=\frac{3}{4} \end{aligned} $$
