Question

If $f\left(x\right)=1+\cos x+{\cos }^{2}x+\cdots \infty$, then $f\left(x\right)+f(\frac{\pi }{2}+x)+f(\frac{\pi }{2}-x)+f(\pi -x)$ equals

• A. ${\sin }^{2}2x{\cos }^{2}2x$
• B. $2{\sec }^{2}x+{\cos }^{2}x$
• C. $2{\sec }^{2}x{\csc }^{2}x$
• D. $2({\sin }^{2}x+{\csc }^{2}x)$

Answer 1

The correct option is C $2{\sec }^{2}x{\csc }^{2}x$

Here $f\left(x\right)=1+\cos x+{\cos }^{2}x+.....\infty$ is in G.P., with common ratio $\cos x$

Hence, $f\left(x\right)=\frac{1}{1-\cos x}$

Now,

$f\left(x\right)+f(\frac{\pi }{2}+x)+f(\frac{\pi }{2}-x)+f(\pi -x)$

$=\frac{1}{1-\cos x}+\frac{1}{1-\cos (\frac{\pi }{2}+x)}+\frac{1}{1-\cos (\frac{\pi }{2}-x)}+\frac{1}{1-\cos (\pi -x)}$

$=\frac{1}{1-\cos x}+\frac{1}{1+\sin x}+\frac{1}{1-\sin x}+\frac{1}{1+\cos x}$

$=(\frac{1}{1-\cos x}+\frac{1}{1+\cos x})+(\frac{1}{1-\sin x}+\frac{1}{1-\sin x})$

$=\frac{2}{{\sin }^{2}x}+\frac{2}{{\cos }^{2}x}$

$=\frac{2}{{\sin }^{2}x{\cos }^{2}x}$

$=2{\sec }^{2}x{\csc }^{2}x$