Question
Prove the following:
$\cos\Bigg(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\Bigg)=\frac{6}{5\sqrt{13}}$.
$\cos\Bigg(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\Bigg)=\frac{6}{5\sqrt{13}}$.
✓
Answer
$\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4},\cot^{-1}\frac{3}{2}=\tan^{-1}\frac{2}{3}$ $ \therefore\cos\Bigg(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3}\Bigg)=\cos\Bigg(\tan^{-1}\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}\Bigg)=\cos\Bigg(\tan^{-1}\frac{17}{6}\Bigg)$= $\cos\Bigg(\cos^{-1}\frac{6}{5\sqrt{13}}\Bigg)=\frac{6}{5\sqrt{13}}=\text{RHS}$.
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