Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.

${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$

$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.

Prove

(a) ${sin}^{-1}\frac{4}{5}+{sin}^{-1}\frac{5}{13}+{sin}^{-1}\frac{16}{65}=\frac{\pi }{2}$

(b) ${sin}^{-1}\frac{3}{5}+{sin}^{-1}\frac{8}{17}={cos}^{-1}\frac{36}{85}$

(c) ${sin}^{-1}\frac{3}{5}+{cos}^{-1}\frac{12}{13}={cos}^{-1}\frac{33}{65}$

Answer 1

To prove ${sin}^{-1}\left(4/5\right)+{sin}^{-1}\left(5/13\right)$

$\frac{\pi }{2}-{sin}^{-1}\frac{16}{65}={cos}^{-1}\left(\frac{16}{65}\right)$.

Now ${sin}^{-1}x\pm {sin}^{-1}y$

$={sin}^{-1}[x\sqrt{1-y^2}\pm y\sqrt{1-x^2}]$.

L.H.S. $={sin}^{-1}[\frac{4}{5}\sqrt{(1-\frac{25}{169})}+\frac{5}{12}\sqrt{(1-\frac{16}{25})}]$

$={sin}^{-1}(\frac{4}{5},\frac{12}{13}+\frac{5}{13},\frac{3}{5})={sin}^{-1}(\frac{48}{65}+\frac{15}{65})$

$={sin}^{-1}\frac{63}{65}$

and R.H.S. $={cos}^{-1}\left(16/65\right)={sin}^{-1}\left(63/65\right)$