# If cos x=

Question:

If $\cos x=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \cos \beta}$, prove that $\tan \frac{x}{2}=\pm \tan \frac{\alpha}{2} \tan \frac{\beta}{2}$

Solution:

Given:

$\cos x=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \cos \beta}$      ....(1)

$\left.\Rightarrow \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \times \cos \beta} \quad \because \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]$

By componendo and dividendo, we get

$\frac{\left(1-\tan ^{2} \frac{x}{2}\right)+\left(1+\tan ^{2} \frac{x}{2}\right)}{\left(1-\tan ^{2} \frac{x}{2}\right)-\left(1+\tan ^{2} \frac{x}{2}\right)}=\frac{(1+\cos \alpha \times \cos \beta+\cos \alpha+\cos \beta)}{-(1+\cos \alpha \cos \beta-\cos \alpha-\cos \beta)}$

$\Rightarrow \frac{2}{2 \tan ^{2} \frac{x}{2}}=\frac{(1+\cos \alpha)(1+\cos \beta)}{(1-\cos \alpha)(1-\cos \beta)}$

$\Rightarrow \tan ^{2} \frac{x}{2}=\frac{(1-\cos \alpha)(1-\cos \beta)}{(1+\cos \alpha)(1+\cos \beta)}$

$\Rightarrow \tan ^{2} \frac{x}{2}=\frac{2 \sin ^{2} \frac{\alpha}{2} \times 2 \sin ^{2} \frac{\beta}{2}}{2 \cos ^{2} \frac{\alpha}{2} \times 2 \cos ^{2} \frac{\beta}{2}}$

$\Rightarrow \tan ^{2} \frac{x}{2}=\tan ^{2} \frac{\alpha}{2} \times \tan ^{2} \frac{\beta}{2}$

$\Rightarrow \tan \frac{x}{2}=\pm \tan \frac{\alpha}{2} \times \tan \frac{\beta}{2}$

Hence proved.