Question

The least value of the expression $\frac{\cot 2x-\tan 2x}{1+\sin (\frac{5\pi }{2}-8x)}$ for $0<x<\frac{\pi }{8}$ is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 2

$\frac{\cot 2x-\tan 2x}{1+\sin (\frac{5\pi }{2}-8x)}$

$=\frac{\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}}{1+\cos \left(8x\right)}$

$=\frac{\frac{2({\cos }^{2}2x-{\sin }^{2}2x)}{2\sin 2x\cos 2x}}{1+\cos \left(8x\right)}$

$=\frac{\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}}{1+\cos \left(8x\right)}$

$=\frac{2\cos 4x}{\sin 4x(1+2{\cos }^{2}4x-1)}$

$=\frac{2\cos 4x}{\sin 4x\left(2{\cos }^{2}4x\right)}=\frac{2}{\sin 8x}=2$

Least value is when $\sin 8x$ is maximum which is equal to $1$.