Question

Prove that sin⁻¹$\frac{3}{5}$ + sin⁻¹$\frac{8}{17}$ = cos⁻¹$\frac{36}{85}$.

Answer 1

Step-by-step explanation:

LHS

As we know that

$sin^{-1}x+sin^{-1}y=sin^{-1}[x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}]\\\\sin^{-1}\frac{3}{5} +sin^{-1}\frac{8}{17}=sin^{-1}[\frac{3}{5}\sqrt{1-(\frac{8}{17})^{2}} +\frac{8}{17}\sqrt{1-(\frac{3}{5})^{2}}]\\\\\\=sin^{-1}[\frac{3}{5}\sqrt{1-(\frac{64}{289})} +\frac{8}{17}\sqrt{1-(\frac{9}{25})}]\\\\\\=sin^{-1}[\frac{3}{5}\sqrt{(\frac{289-64}{289})} +\frac{8}{17}\sqrt{(\frac{16}{25})}]\\\\\\=sin^{-1}[\frac{3}{5}\sqrt{(\frac{225}{289})} +\frac{8}{17}\sqrt{(\frac{16}{25})}]$

$=sin^{-1}[\frac{45}{85}+\frac{32}{85}]\\\\\\=sin^{-1}[\frac{77}{85}]$....eq1

RHS

$cos^{-1}x=sin^{-1}\sqrt{1-x^{2}} \\\\cos^{-1}\frac{36}{85} =sin^{-1}\sqrt{1-(\frac{36}{85})^{2}} \\\\\\=sin^{-1}\sqrt{(\frac{7225-1296}{7225})} \\\\\\=sin^{-1}\frac{77}{85}$...eq2

from eq1 and eq2

LHS=RHS

hence proved