\( \sin \left(\tan ^{-1} x\right),|x|<1  \\ \because \tan ^{-1} x   =\sin ^{-1} \frac{x}{\sqrt{1+x^2}} \\ \therefore \sin \left(\tan ^{-1} x\right)   =\sin \left[\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right]=\frac{x}{\sqrt{1+x^2}} \)

 \(\alpha=\cot ^{-1} \frac{1}{2} \) என்க \(\Rightarrow \cot \alpha=\frac{1}{2}\) மற்றும்\(\ \cos \alpha=\frac{1}{\sqrt{5}} \)
\(\therefore \sin ^{-1}(x)+\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{\pi}{2} \\ \Rightarrow x=\frac{1}{\sqrt{5}} \quad\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)\)

\( \sin ^{-1}\left(2 \cos ^2 x-1\right)+\cos ^{-1}\left(1-2 \sin ^2 x\right) \\ =\sin ^{-1}(\cos 2 x)+\cos ^{-1}(\cos 2 x) \\ =\frac{\pi}{2} \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] \)

 \(\cot ^{-1} 2+\cot ^{-1} 3 =\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3} \\   =\tan ^{-1}\left[\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{b}{6}}\right]=\tan ^{-1}(1)=\frac{\pi}{4}\)
மூன்றாவது கோணம் = \(\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \)

\( \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \cdot \frac{2}{9}}\right) \\ =\tan ^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right)=\tan ^{-1}\left(\frac{17}{34}\right)=\tan ^{-1}\left(\frac{1}{2}\right) \)

\( \tan ^{-1} x+\cos ^{-1} x =\frac{\pi}{2} \\ \tan ^{-1} x =\frac{\pi}{2}-\cot ^{-1} x=\frac{\pi}{2}-\frac{2 \pi}{5} \\ =\frac{5 \pi-4 \pi}{10}=\frac{\pi}{10} \)

2sin⁻¹ \(\alpha\) -க்கு ஒரு தீர்வு இருந்தால்
\( -\frac{\pi}{2} \leq 2 \sin ^{-1} \alpha \leq \frac{\pi}{2} \\ -\frac{\pi}{4} \leq \sin ^{-1} \alpha \leq \frac{\pi}{4} \\ \sin \left(\frac{-\pi}{4}\right) \leq \alpha \leq \sin \frac{\pi}{4} \\ -\frac{1}{\sqrt{2}} \leq \alpha \leq \frac{1}{\sqrt{2}} \\ |\alpha| \leq \frac{1}{\sqrt{2}} \)

\(\sin ^{-1}(\cos x)=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-x\right)\right)=\frac{\pi}{2}-x\)