Question

Prove that ${\sin }^{-1}\frac{3}{5}-{\sin }^{-1}\frac{8}{17}={\cos }^{-1}\left(\frac{84}{85}\right)$

Answer 1

L.H.S

${\sin }^{-1}\frac{3}{5}-{\sin }^{-1}\frac{8}{12}$

We know that

${\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}\{x\sqrt{1-y^2}-y\sqrt{1-x^2}\}$

Therefore,

$={\sin }^{-1}\{\frac{3}{5}\sqrt{1-{\left(\frac{8}{17}\right)}^2}-\frac{8}{17}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\}$

$={\sin }^{-1}\{\frac{3}{5}\sqrt{1-\frac{64}{289}}-\frac{8}{17}\sqrt{1-\frac{9}{25}}\}$

$={\sin }^{-1}\{\frac{3}{5}\sqrt{\frac{225}{289}}-\frac{8}{17}\sqrt{\frac{16}{25}}\}$

$={\sin }^{-1}\{\frac{3}{5}\times \frac{15}{17}-\frac{8}{17}\times \frac{4}{5}\}$

$={\sin }^{-1}\{\frac{45}{85}-\frac{32}{85}\}$

$={\sin }^{-1}\left\{\frac{13}{85}\right\}$

$={\cos }^{-1}\left(\frac{84}{85}\right)$

Hence, proved.