Prove that

Question:

Prove that $2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$

Solution:

$\mathrm{LHS}=2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)$ $=\cos ^{-1}\left\{\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right\}$   $\left[\because 2 \tan ^{-1}(x)=\cos ^{-1}\left\{\frac{1-x^{2}}{1+x^{2}}\right\}\right]$

$=\cos ^{-1}\left\{\frac{1-\frac{a-b}{a+b} \tan ^{2} \frac{\theta}{2}}{1+\frac{a-b}{a+b} \tan ^{2} \frac{\theta}{2}}\right\}$

$=\cos ^{-1}\left\{\frac{a+b-(a-b) \tan ^{2} \frac{\theta}{2}}{a+b+(a-b) \tan ^{2} \frac{\theta}{2}}\right\}$

$=\cos ^{-1}\left\{\frac{a+b-a \tan ^{2} \frac{\theta}{2}+b \tan ^{2} \frac{\theta}{2}}{a+b+a \tan ^{2} \frac{\theta}{2}-b \tan ^{2} \frac{\theta}{2}}\right\}$

$=\cos ^{-1}\left\{\frac{a\left(1-\tan ^{2} \frac{\theta}{2}\right)+b\left(1+\tan ^{2} \frac{\theta}{2}\right)}{a\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right\}$

$=\cos ^{-1}\left\{\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b\left(\frac{1+\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)}{a\left(\frac{1+\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}\right)}\right\}$                 [Dividing $N^{r}$ and $D^{r}$ by $1+\tan ^{2} \frac{\theta}{2}$ ]

$=\cos ^{-1}\left\{\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}\right)}\right\}$

$=\cos ^{-1}\left\{\frac{a \cos \theta+b}{a+b \cos \theta}\right\}=$ RHS

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