Question

(i) Evaluate ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$
(ii) Show that: ${\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\frac{\pi }{4}-\frac{x}{2}$

Answer 1

(i)

${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=2n\pi \pm \frac{\pi }{3}$ .... (general solution)

(ii)

$\frac{\cos x}{1-\sin x}=\frac{\cos 2\left(\frac{x}{2}\right)}{1-\sin 2\left(\frac{x}{2}\right)}$

$=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\cos \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)}$

$=\frac{\{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\left\}\right\{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\}}{{\{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\}}^2}$

$=\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}$

Dividing numerator and denominator by $\cos \left(\frac{x}{2}\right)$ we get

$\frac{\cos x}{1-\sin x}=\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\tan (\frac{\pi }{4}-\frac{x}{2})$

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

Hence, ${\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\frac{\pi }{4}-\frac{x}{2}$