To prove: $\sin^{-1}\left(\frac{4}{5}\right)+\sin^{-1}\left(\frac{5}{13}\right)$ + $\sin^{-1}\left(\frac{16}{65}\right)$ = $\frac{\pi}{2}$ Let $\sin^{-1}\left(\frac{4}{5}\right)$ = x ⇒ sin x = $\frac{4}{5}$ ⇒ cos x = $\sqrt{1-\sin^2 x}$ = $\frac{3}{5}$ $\sin^{-1}\left(\frac{5}{13}\right)$ = y ⇒ sin y = $\frac{5}{13}$ ⇒ cos y = $\sqrt{1-\sin^2 y}$ = $\frac{12}{13}$ $\sin^{-1}\left(\frac{16}{65}\right)$ = z ⇒ sin z = $\frac{16}{65}$ ⇒ cos z = $\sqrt{1-\sin^2 x}$ = $\frac{63}{65}$ tan x = $\frac{4}{3}$ , tan y = $\frac{5}{12}$ , tan z = $\frac{16}{63}$ tan z = $\frac{16}{63}$ ⇒ cot z = $\frac{63}{16}$ ... (1) tan (x + y) = $\frac{\tan(x+y)}{1-\tan x \tan y}$ ⇒ tan (x + y) = $\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}$ ⇒ tan (x + y) = $\frac{63}{16}$ ⇒ tan (x + y) = cot z ... [from equation (1)] ⇒ tan (x + y) = tan $\left(\frac{\pi}{2}-z\right)$ ⇒ x + y = $\frac{\pi}{2}$ - z ⇒ x + y + z = $\frac{\pi}{2}$ ∴ $\sin^{-1}\left(\frac{4}{5}\right)+\sin^{-1}\left(\frac{5}{13}\right)$ + $\sin^{-1}\left(\frac{16}{65}\right)$ = $\frac{\pi}{2}$ OR $\tan^{-1}3x+\tan^{-1}2x$ = $\frac{\pi}{4}$ ⇒ $\tan^{-1}\left(\frac{5x}{1-6x^2}\right)$ = $\frac{\pi}{4}$ . 3x × 2x < 1 ⇒ tan $\left[\tan^{-1}\left(\frac{5x}{1-6x^2}\right)\right]$ = tan $\frac{\pi}{4}$ ⇒ $\frac{5x}{1-6x^2}$ = 1 ⇒ 1 - $6x^2$ = 5x ⇒ $6x^2$ + 5x - 1 = 0 ⇒ $6x^2$ + 6x - x - 1 = 0 ⇒ x = - 1 or $\frac{1}{6}$ Here (- 3) × (- 2) ≮ 1 [Since (- 3) × (- 2) = 6 >1] Therefore, x = 1 is not the solution. When substituting x = $\frac{1}{6}$ in 3x 2x, we have, 3 × $\frac{1}{6}$ × 2 × $\frac{1}{6}$ = $\frac{1}{2}\times\frac{1}{3}$ = $\frac{1}{6}$ < 1 Hence x = $\frac{1}{6}$ is the solution of the given equation