Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) Prove that $2{tan}^2\left\{tan\frac{\alpha }{2}tan(\frac{\pi }{4}-\frac{\beta }{2})\right\}={tan}^{-1}\left[\frac{sin\alpha cos\beta }{sin\beta +cos\alpha }\right]$
(b) Prove that
${cos}^{-1}\left(\frac{2+3cosx}{3+2cosx}\right)=2{tan}^{-1}\left(\frac{1}{\sqrt{5}}tan\frac{x}{2}\right)$.

Answer 1

(a) We know that $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$
$\therefore$ L.H.S.${\tan }^{-1}\frac{2\tan \left(\alpha /2\right)\tan (\pi /4-\beta /2)}{1-{\tan }^2\left(\alpha /2\right){\tan }^2(\pi /4-\beta /2)}$
Now change in terms of \sin e and \cos
$2\sin \left(\alpha /2\right)\cos \left(\alpha /2\right)$
$={\tan }^{-1}\frac{\sin (\pi /4-\beta /2)\cos (\pi /4-\beta /2)}{{\cos }^2\left(\alpha /2\right){\cos }^2(\pi /4-\beta /2)}-{\sin }^2\left(\alpha /2\right){\sin }^2(\pi /4-\beta /2)$
$={\tan }^{-1}\frac{\sin \alpha .\sin (\pi /2-\beta )}{2\left[\cos \right(\alpha /2-\beta /2+\pi /4\left)\cos \right(\alpha /2+\beta /2-\pi /4\left)\right]}$
$={\tan }^{-1}\frac{\sin \alpha \cos \beta }{\cos \alpha +\cos (\pi /2-\beta )}$
$={\tan }^{-1}\frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta }$
(b) Let $\theta ={\cos }^{-1}\frac{2+3\cos x}{3+2\cos x}$
$\therefore$ $\frac{\cos \theta }{1}=\frac{2+3\cos x}{3+2\cos x}$ [Apply Compo. & Divi.]
$\frac{1+\cos \theta }{1+\cos \theta }=\frac{1-\cos x}{5(1+\cos x)}$
$\therefore$ ${\tan }^2\frac{\theta }{2}=\frac{1}{5}{\tan }^2\frac{x}{2}$
$\therefore$ $\theta =2{\tan }^{-1}\left[\frac{1}{\sqrt{5}}\tan \frac{x}{2}\right]=$ R.H.S.