Question

If $\cos \theta =\frac{2\cos \varphi -1}{2-\cos \varphi }$, prove that $\tan \left(\theta /2\right)=\sqrt{3}\tan \left(\varphi /2\right)$ and hence show that $\sin \varphi =\frac{\sqrt{3}\sin \theta }{2+\cos \theta }$

Answer 1

Given, $\cos \varphi =\frac{2\cos \varphi -1}{2-\cos \varphi }$;

To prove (i) $\tan \theta /2=\sqrt{3}\tan \varphi /2$

(ii) $\sin \varphi =\frac{\sqrt{3}\sin \theta }{2+\cos \theta }$

Proof: (i) We know, $\cos A=\frac{1-{\tan }^{2}A/2}{1+{\tan }^{2}A/2}$

$\Rightarrow \frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}=\frac{2\left[\frac{1-{\tan }^2\varphi /2}{1+{\tan }^2\varphi /2}\right]-1}{2-\left[\frac{1-{\tan }^2\varphi /2}{1+{\tan }^2\varphi /2}\right]}$

$\Rightarrow \frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}=\frac{2-2{\tan }^2\varphi /2-1-{\tan }^2\varphi /2}{2+2{\tan }^2\varphi /2-1+{\tan }^2\varphi /2}$

$=\frac{1-3{\tan }^2\varphi /2}{1+3{\tan }^2\varphi /2}$

Applying componendo and dividendo,

$\frac{2}{-2{\tan }^2\theta /2}=\frac{2}{-6{\tan }^2\varphi /2}$

$\Rightarrow {\tan }^2\frac{\theta }{2}=3{\tan }^2\frac{\varphi }{2}$

$\Rightarrow \tan \frac{\theta }{2}=\sqrt{3}\tan \frac{\varphi }{2}$ proved.

(ii) $\sin \varphi =\frac{\sqrt{3}\sin \theta }{2+\cos \theta }$

We know $\tan \frac{\theta }{2}=\sqrt{3}\tan \frac{\varphi }{2}$

$\Rightarrow \tan \frac{\varphi }{2}=\frac{1}{\sqrt{3}}\tan \frac{\theta }{2}$

$\Rightarrow \frac{2\tan \varphi /2}{1+{\tan }^2\varphi /2}=\frac{\frac{2}{\sqrt{3}}\tan \theta /2}{1+\frac{1}{3}{\tan }^2\frac{\theta }{2}}$

$\Rightarrow \sin \varphi =\frac{\frac{2}{\sqrt{3}}\tan \frac{\theta }{2}}{1+\frac{1}{3}{\tan }^2\theta /2}$

$=\frac{2\sqrt{3}\tan \frac{\theta }{2}}{3+{\tan }^2\frac{\theta }{2}}$

$=\frac{\sqrt{3}\frac{2\tan \theta /2}{1+{\tan }^2\theta /2}}{\frac{2+2{\tan }^2\theta /2+1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}}$

$=\frac{\sqrt{3}\sin \theta }{2+\cos \theta }$= R.H.S proved .