$\left[\sin \left( an ^{-1} \frac{3}{4} ight) ight]^{2}+\left[\sin \left( an ^{-1} \frac{4}{3} ight) ight]^{2}=$

Question

$\left[\sin \left(	an ^{-1} \frac{3}{4}ight)ight]^{2}+\left[\sin \left(	an ^{-1} \frac{4}{3}ight)ight]^{2}=$

Aicharya · Accepted Answer

Let $	an ^{-1} \frac{3}{4}=	heta \Rightarrow 	an 	heta=\frac{3}{4} \Rightarrow \cot 	heta=\frac{4}{3}$ $	herefore \operatorname{cosec}^{2} 	heta=1+\frac{16}{9}=\frac{25}{9} \Rightarrow \operatorname{cosec} 	heta=\frac{5}{3} \Rightarrow \sin 	heta=\frac{3}{5}$ $	herefore \sin \left(	an ^{-1} \frac{3}{4}ight)=\sin \left(\sin ^{-1} \frac{3}{5}ight)=\frac{3}{5}$ Let $	an ^{-1} \frac{4}{3}=\phi \Rightarrow 	an \phi=\frac{4}{3} \Rightarrow \cot \phi=\frac{3}{4}$ $	herefore \operatorname{cosec}^{2} \phi=1+\frac{9}{16}=\frac{25}{16} \Rightarrow \operatorname{cosec} \phi=\frac{5}{4} \Rightarrow \sin \phi=\frac{4}{5}$ $	herefore \sin \left(	an ^{-1} \frac{4}{3}ight)=\sin \left(\sin ^{-1} \frac{4}{5}ight)=\frac{4}{5}$ Hence given expression $=\left(\frac{3}{5}ight)^{2}+\left(\frac{4}{5}ight)^{2}=\frac{9+16}{25}=1$

$\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^{2}+\left[\sin \left(\tan ^{-1}…

Solution