Question

Prove that: ${\tan }^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)=\frac{\pi }{4}-\frac{x}{2}$, where $\pi <x<\frac{3\pi }{2}$

Answer 1

To prove : ${\tan }^{-1}\left\{\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}\right\}=\frac{\pi }{4}-\frac{x}{2}$ if $xϵ(\pi ,\frac{3\pi }{2})$

Proof : We know that ; $1+cosx=2{\cos }^2\frac{x}{2}$

and, $1-cosx=2{\sin }^2\frac{x}{2}$

From $L.H.S$

${\tan }^{-1}\left\{\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}\right\}={\tan }^{-1}\left\{\frac{\sqrt{2{\cos }^2\frac{x}{2}}+\sqrt{2{\sin }^2\frac{x}{2}}}{\sqrt{2{\cos }^2\frac{x}{2}}-\sqrt{2{\sin }^2\frac{x}{2}}}\right\}$

$={\tan }^{-1}\left\{\frac{\sqrt{2}\left|cosx/2\right|+\sqrt{2}\left|sinx/2\right|}{\sqrt{2}\left|cosx/2\right|-\sqrt{2}\left|sinx/2\right|}\right\}$

now, we have, $\pi <x<3\pi /2\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$

i.e. in the second quadrant where $\left|cosx\right|=-cosx$ and $\left|sinx\right|=+sinx$

$\therefore$ $={\tan }^{-1}\left\{\frac{-cosx/2+sinx/2}{-cosx/2-sinx/2}\right\}$

Now, divide the numerator and denominator by $-cosx/2$

$={\tan }^{-1}\left\{\frac{1-tanx/2}{1+tanx/2}\right\}$

$={tan}^{-1}\left\{\frac{\tan \pi /4-tanx/2}{1+\tan \frac{\pi }{4}.tanx/2}\right\}$ ($\because$ $tan\pi /4=1$)

$={\tan }^{-1}\left\{\tan (\frac{\pi }{4}-\frac{x}{2})\right\}$ (Using $\tan (A-B)=\frac{tanA-tanB}{1+tanA.tanB}$)

Now, we know that

${\tan }^{-1}\left(\tan \theta \right)=\theta$ if $\theta ϵ(-\pi /2,\pi /2)$

We have from question, that

$\pi <x<3\pi /2$

$\Rightarrow -3\pi /2<-x<-\pi$

$\Rightarrow -3\pi /4<-x/2<-\pi /2\Rightarrow -\pi /2<(\pi /4-x/2)<-\pi /4$

So basically $(\pi /4-x/2)ϵ(-\pi /2,-\pi /4)$ which lies with in $(-\pi /2,\pi /2)$

$\therefore$ ${tan}^{-1}\left\{tan(\frac{\pi }{4}-\frac{x}{2})\right\}=\frac{\pi }{4}-\frac{x}{2}$

Hence proved.