Question

If $\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta },(0<\alpha <\pi ,0<\beta <\pi )$ , then $\tan \frac{\alpha }{2}\cot \frac{\beta }{2}$ is equal to

• A. $1$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. none of these

Answer 1

The correct option is C $\sqrt{3}$

Given

$\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta }$

Applying Compoundo and Dividendo rule

$\frac{1+\cos \alpha }{1-\cos \alpha }=\frac{2\cos \beta -1+2-\cos \beta }{2-\cos \beta -2\cos \beta +1}$

$\frac{2{\cos }^2\frac{\alpha }{2}}{2{\sin }^2\frac{\alpha }{2}}=\frac{1+\cos \beta }{3(1-\cos \beta )}$

${\cot }^2\frac{\alpha }{2}=\frac{2{\cos }^2\frac{\beta }{2}}{3\left(2{\sin }^2\frac{\beta }{2}\right)}$

$⟹{\cot }^2\frac{\alpha }{2}=\frac{1}{3}{\cot }^2\frac{\beta }{2}$

$⟹{\tan }^2\frac{\alpha }{2}{\cot }^2\frac{\beta }{2}=3$

$⟹\tan \frac{\alpha }{2}\cot \frac{\beta }{2}=\sqrt{3}$