Question:
If \(\alpha = \cos^{-1}\left(\frac{3}{5}\right)\), \(\beta = \tan^{-1}\left(\frac{1}{3}\right)\), where \(0 \lt \alpha, \beta \lt \frac{\pi}{2}\), then \(\alpha - \beta\) is equal to :
If \(\alpha = \cos^{-1}\left(\frac{3}{5}\right)\), \(\beta = \tan^{-1}\left(\frac{1}{3}\right)\), where \(0 \lt \alpha, \beta \lt \frac{\pi}{2}\), then \(\alpha - \beta\) is equal to :
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Find \(\sin\alpha, \cos\beta, \sin\beta\) from the right triangles and use \(\sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\).
Updated On: Nov 5, 2026
- \(\tan^{-1}\left(\frac{9}{14}\right)\)
- \(\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)\)
- \(\cos^{-1}\left(\frac{9}{5\sqrt{10}}\right)\)
- \(\tan^{-1}\left(\frac{9}{5\sqrt{10}}\right)\)
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The Correct Option is B
Solution and Explanation
\(\cos\alpha = \frac{3}{5} \Rightarrow \sin\alpha = \frac{4}{5}\). \(\tan\beta = \frac{1}{3} \Rightarrow \sin\beta = \frac{1}{\sqrt{10}},\ \cos\beta = \frac{3}{\sqrt{10}}\).
\(\sin(\alpha-\beta) = \frac{4}{5}\cdot\frac{3}{\sqrt{10}} - \frac{3}{5}\cdot\frac{1}{\sqrt{10}} = \frac{12-3}{5\sqrt{10}} = \frac{9}{5\sqrt{10}}\).
Since \(\alpha \approx 53.1^\circ\) and \(\beta \approx 18.4^\circ\), \(\alpha-\beta \approx 34.7^\circ\) lies in \((0, \frac{\pi}{2})\), so \(\alpha - \beta = \sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)\).
Answer: (2) \(\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)\)
\(\sin(\alpha-\beta) = \frac{4}{5}\cdot\frac{3}{\sqrt{10}} - \frac{3}{5}\cdot\frac{1}{\sqrt{10}} = \frac{12-3}{5\sqrt{10}} = \frac{9}{5\sqrt{10}}\).
Since \(\alpha \approx 53.1^\circ\) and \(\beta \approx 18.4^\circ\), \(\alpha-\beta \approx 34.7^\circ\) lies in \((0, \frac{\pi}{2})\), so \(\alpha - \beta = \sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)\).
Answer: (2) \(\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)\)
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