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

(D)
Let \(\tan ^{-1} \frac{3}{4}=\theta \Rightarrow \tan \theta=\frac{3}{4} \Rightarrow \cot \theta=\frac{4}{3}\)
\(\therefore \operatorname{cosec}^{2} \theta=1+\frac{16}{9}=\frac{25}{9} \Rightarrow \operatorname{cosec} \theta=\frac{5}{3} \Rightarrow \sin \theta=\frac{3}{5}\)
\(\therefore \sin \left(\tan ^{-1} \frac{3}{4}\right)=\sin \left(\sin ^{-1} \frac{3}{5}\right)=\frac{3}{5}\)
Let \(\tan ^{-1} \frac{4}{3}=\phi \Rightarrow \tan \phi=\frac{4}{3} \Rightarrow \cot \phi=\frac{3}{4}\)
\(\therefore \operatorname{cosec}^{2} \phi=1+\frac{9}{16}=\frac{25}{16} \Rightarrow \operatorname{cosec} \phi=\frac{5}{4} \Rightarrow \sin \phi=\frac{4}{5}\)
\(\therefore \sin \left(\tan ^{-1} \frac{4}{3}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)
Hence given expression \(=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=\frac{9+16}{25}=1\)