Solveeit Logo
Login

Question

Question: If \(\tan\beta = \cos\theta\tan\alpha,\) then \(\tan^{2}\frac{\theta}{2} =\)...

If tan⁡β=cos⁡θtan⁡α,\tan\beta = \cos\theta\tan\alpha, then tan⁡2θ2=\tan^{2}\frac{\theta}{2} =

A

sin⁡(α+β)sin⁡(α−β)\frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)}

B

cos⁡(α−β)cos⁡(α+β)\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)}

C

sin⁡(α−β)sin⁡(α+β)\frac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)}

D

cos⁡(α+β)cos⁡(α−β)\frac{\cos(\alpha + \beta)}{\cos(\alpha - \beta)}

Answer

sin⁡(α−β)sin⁡(α+β)\frac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)}

Explanation

Solution

tan⁡2θ2=1−cos⁡θ1+cos⁡θ=tan⁡α−tan⁡βtan⁡α+tan⁡β=sin⁡(α−β)sin⁡(α+β)\tan^{2}\frac{\theta}{2} = \frac{1 - \cos\theta}{1 + \cos\theta} = \frac{\tan\alpha - \tan\beta}{\tan\alpha + \tan\beta} = \frac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)}.