1. **Problem statement:** Prove that $$\sin^{-1} \frac{3}{5} + \frac{1}{2} \cos^{-1} \frac{5}{13} - \cot^{-1} 2 = \tan^{-1} \frac{28}{29}$$. 2. **Recall formulas and identities:** - Inverse sine and cosine are related by $$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$. - Half-angle formulas and sum/difference formulas for inverse trigonometric functions. - Relationship between $$\cot^{-1} x$$ and $$\tan^{-1} x$$: $$\cot^{-1} x = \tan^{-1} \frac{1}{x}$$. 3. **Evaluate each term:** - Let $$A = \sin^{-1} \frac{3}{5}$$. - Let $$B = \cos^{-1} \frac{5}{13}$$. 4. **Calculate $$\sin A$$ and $$\cos A$$:** - Given $$\sin A = \frac{3}{5}$$, so $$\cos A = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$$. 5. **Calculate $$\cos B$$ and $$\sin B$$:** - Given $$\cos B = \frac{5}{13}$$, so $$\sin B = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$$. 6. **Express $$\frac{1}{2} B$$ as $$\frac{1}{2} \cos^{-1} \frac{5}{13}$$:** - Use half-angle formula for cosine: $$\cos \frac{B}{2} = \sqrt{\frac{1 + \cos B}{2}} = \sqrt{\frac{1 + \frac{5}{13}}{2}} = \sqrt{\frac{\frac{18}{13}}{2}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}$$. - Use half-angle formula for sine: $$\sin \frac{B}{2} = \sqrt{\frac{1 - \cos B}{2}} = \sqrt{\frac{1 - \frac{5}{13}}{2}} = \sqrt{\frac{\frac{8}{13}}{2}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$$. 7. **Calculate $$\sin \left( A + \frac{B}{2} \right)$$ and $$\cos \left( A + \frac{B}{2} \right)$$:** - Using sum formulas: $$\sin \left( A + \frac{B}{2} \right) = \sin A \cos \frac{B}{2} + \cos A \sin \frac{B}{2} = \frac{3}{5} \cdot \frac{3}{\sqrt{13}} + \frac{4}{5} \cdot \frac{2}{\sqrt{13}} = \frac{9}{5\sqrt{13}} + \frac{8}{5\sqrt{13}} = \frac{17}{5\sqrt{13}}$$. $$\cos \left( A + \frac{B}{2} \right) = \cos A \cos \frac{B}{2} - \sin A \sin \frac{B}{2} = \frac{4}{5} \cdot \frac{3}{\sqrt{13}} - \frac{3}{5} \cdot \frac{2}{\sqrt{13}} = \frac{12}{5\sqrt{13}} - \frac{6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}}$$. 8. **Calculate $$\tan \left( A + \frac{B}{2} \right)$$:** $$\tan \left( A + \frac{B}{2} \right) = \frac{\sin \left( A + \frac{B}{2} \right)}{\cos \left( A + \frac{B}{2} \right)} = \frac{\frac{17}{5\sqrt{13}}}{\frac{6}{5\sqrt{13}}} = \frac{17}{6}$$. 9. **Rewrite the original expression:** $$\sin^{-1} \frac{3}{5} + \frac{1}{2} \cos^{-1} \frac{5}{13} - \cot^{-1} 2 = \left( A + \frac{B}{2} \right) - \cot^{-1} 2$$. 10. **Express $$\cot^{-1} 2$$ as $$\tan^{-1} \frac{1}{2}$$:** $$\cot^{-1} 2 = \tan^{-1} \frac{1}{2}$$. 11. **Use the formula for difference of arctangents:** $$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x - y}{1 + xy}$$. 12. **Apply to our values:** $$x = \frac{17}{6}, y = \frac{1}{2}$$ $$\tan^{-1} \frac{17}{6} - \tan^{-1} \frac{1}{2} = \tan^{-1} \frac{\frac{17}{6} - \frac{1}{2}}{1 + \frac{17}{6} \cdot \frac{1}{2}} = \tan^{-1} \frac{\frac{17}{6} - \frac{3}{6}}{1 + \frac{17}{12}} = \tan^{-1} \frac{\frac{14}{6}}{\frac{29}{12}} = \tan^{-1} \frac{14}{6} \cdot \frac{12}{29} = \tan^{-1} \frac{168}{174} = \tan^{-1} \frac{28}{29}$$. 13. **Conclusion:** We have shown that $$\sin^{-1} \frac{3}{5} + \frac{1}{2} \cos^{-1} \frac{5}{13} - \cot^{-1} 2 = \tan^{-1} \frac{28}{29}$$ which completes the proof.