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Question: If \(\tan\beta = \cos\theta.\tan\alpha\) then \(\tan^{2}\theta/2\)equal to...

If tan⁡β=cos⁡θ.tan⁡α\tan\beta = \cos\theta.\tan\alpha then tan⁡2θ/2\tan^{2}\theta/2equal to

A

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

B

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

C

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

D

cos⁡(α+β)/cos⁡(α−β)\cos(\alpha + \beta)/\cos(\alpha - \beta)

Answer

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

Explanation

Solution

The given relation is tan⁡αtan⁡β=1cos⁡θ\frac{\tan\alpha}{\tan\beta} = \frac{1}{\cos\theta}

Applying componendo and dividendo rule, then

⇒ tan⁡α−tan⁡βtan⁡α+tan⁡β=1−cos⁡θ1+cos⁡θ\frac{\tan\alpha - \tan\beta}{\tan\alpha + \tan\beta} = \frac{1 - \cos\theta}{1 + \cos\theta} ⇒ sin⁡(α−β)sin⁡(α+β)=2sin⁡2θ22cos⁡2θ2\frac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)} = \frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}

⇒ sin⁡(α−β)sin⁡(α+β)=tan⁡2θ2\frac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)} = \tan^{2}\frac{\theta}{2}.