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Question:

If $\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right)$, where $0<\alpha, \beta<\frac{\pi}{2}$, then $\alpha-\beta$ is equal to :

1. (1) $\tan ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

2. (2) $\cos ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

3. (3) $\tan ^{-1}\left(\frac{9}{14}\right)$

4. (4) $\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

Correct Option: , 4

Solution:

$\because \cos \alpha=\frac{3}{5}$, then $\sin \alpha=\frac{4}{5}$

$\Rightarrow \tan \alpha=\frac{4}{3}$

and $\tan \beta=\frac{1}{3}$

$\because \tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}$

$=\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}=\frac{1}{\frac{13}{9}}=\frac{9}{13}$

$\therefore \alpha-\beta=\tan ^{-1}\left(\frac{9}{13}\right)=\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

$=\cos ^{-1}\left(\frac{13}{5 \sqrt{10}}\right)$