$\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ = $\tan^{-1}$ $\left[\frac{\sin(\pi/2 - x)}{1+\cos(\pi/2 - x)}\right]$ = [Since sin θ = 2 sin (θ/2) cos (θ/2) and 1 + cos θ = 2 $\cos^2$ (θ/2)] $\tan^{-1}$ $\left[\tan(\pi/4 - x/2)\right]$ = $(\pi/4 - x/2)$ (proved) OR Let $\sin^{-1}\left(\frac{8}{17}\right)$ = x Then , sin x = $\frac{8}{17}$ ; cos x = $\sqrt{1-x^2}$ ⇒ cos x = $\sqrt{1-\left(\frac{8}{17}\right)^2}$ ⇒ cos x = $\sqrt{1-\left(\frac{8}{17}\right)^2}$ ⇒ cos x = $\sqrt{\frac{225}{289}}$ ⇒ cos x = $\frac{15}{17}$ ∴ tan x = $\frac{\sin x}{\cos x}$ ⇒ tan x = $\frac{\frac{8}{17}}{\frac{15}{17}}$ ⇒ tan x = $\frac{8}{15}$ ⇒ x = $\tan^{-1}\left(\frac{8}{15}\right)$ ... (1) Let $\sin^{-1}\left(\frac{3}{5}\right)$ = y ... (2) Then, sin y = $\frac{3}{5}$ ; cos y = $\sqrt{1-y^2}$ ⇒ cos y = $\sqrt{1-\left(\frac{3}{5}\right)^2}$ ⇒ cos y = $\sqrt{\frac{16}{25}}$ ⇒ cos y = $\frac{4}{5}$ ∴ tan y = $\frac{\sin y}{\cos y}$ ⇒ tan y = $\frac{\frac{3}{5}}{\frac{4}{5}}$ ⇒ tan y = $\frac{3}{4}$ ⇒ y = $\tan^{-1}\left(\frac{3}{4}\right)$ ... (3) From equations (2) and (3), we have, $\sin^{-1}\left(\frac{3}{5}\right)$ = $\tan^{-1}\left(\frac{3}{4}\right)$ Now consider $\sin^{-1}\left(\frac{8}{17}\right)$ + $\sin^{-1}\left(\frac{3}{5}\right)$ From equations (1) and (3), we have, $\sin^{-1}\left(\frac{8}{17}\right)$ + $\sin^{-1}\left(\frac{3}{5}\right)$ = $\tan^{-1}\left(\frac{5}{15}\right) + \tan^{-1}\left(\frac{3}{4}\right)$ = $\tan^{-1}\left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times\frac{3}{4}}\right)$ [Since $\tan^{-1}x + \tan^{-1}y$ = $\tan^{-1}\left(\frac{x+y}{x-y}\right)$] = $\tan^{-1}\left(\frac{32+45}{60-24}\right)$ $\sin^{-1}\left(\frac{8}{17}\right)$ + $\sin^{-1}\left(\frac{3}{5}\right)$ = $\tan^{-1}\left(\frac{77}{36}\right)$ ... (4) Now, we have: Let $\tan^{-1}\left(\frac{77}{36}\right)$ = z Then tan z = $\frac{77}{36}$ ⇒ sec z = $\sqrt{1+\left(\frac{77}{36}\right)^2}$ [Since sec θ = $\sqrt{1+\tan^2\theta}$] ⇒ sec z = $\sqrt{\frac{1296+5929}{1296}}$ ⇒ sec z = $\sqrt{\frac{7225}{1296}}$ ⇒ sec z = $\frac{85}{36}$ We know that cosz = $\frac{1}{\sec z}$ Thus, sec z = $\frac{85}{36}$ , cos z = $\frac{36}{85}$ ⇒ z = $\cos^{-1}\left(\frac{36}{85}\right)$ ⇒ $\tan^{-1}\left(\frac{77}{36}\right)$ = $\cos^{-1}\left(\frac{36}{85}\right)$ ⇒ $\sin^{-1}\left(\frac{8}{17}\right)$ + $\sin^{-1}\left(\frac{3}{5}\right)$ = $\cos^{-1}\left(\frac{36}{85}\right)$ [Since from equation (4)] Hence proved.