Question

If $\cos \alpha +\cos \beta =a,\sin \alpha +\sin \beta =b$ and $\alpha -\beta =2\theta$ then $\frac{\cos 3\theta }{\cos \theta }$ is equal to

• A. $a^2+b^2-2$
• B. $a^2+b^2-3$
• C. $3-a^2-b^2$
• D. $\frac{(a^2+b^2)}{4}$

Answer 1

Answer: (B)

$a^2+b^2-3$

Given $\cos \alpha +\cos \beta =a$ ....(1)
$\sin \alpha +\sin \beta =b$ ....(2)
Squaring and adding (1) and (2), we get
$\Rightarrow 2+2\cos (\alpha -\beta )=a^2+b^2$
$\Rightarrow 4{\cos }^2\theta =a^2+b^2$ ....(3)
Consider, $\frac{\cos 3\theta }{\cos \theta }$
$=\frac{4{\cos }^3\theta -3\cos \theta }{\cos \theta }$
$=a^2+b^2-3$ (using 3)