JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
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
- A \({\sin ^{ - 1}}\,\left( {\frac{9}{{5\sqrt {10} }}} \right)\)
- B \({\cos ^{ - 1}}\,\left( {\frac{9}{{5\sqrt {10} }}} \right)\)
- C \({\tan ^{ - 1}}\,\left( {\frac{9}{{5\sqrt {10} }}} \right)\)
- D \({\tan ^{ - 1}}\,\left( {\frac{9}{{14}}} \right)\)
Answer & Solution
Correct Answer
(A) \({\sin ^{ - 1}}\,\left( {\frac{9}{{5\sqrt {10} }}} \right)\)
Step-by-step Solution
Detailed explanation
\(\cos \alpha = \frac{3}{5},\tan \beta \frac{1}{3}\) \( \Rightarrow \tan \alpha = \frac{4}{3}\) \( \Rightarrow \tan \left( {\alpha - \beta } \right) = \frac{{\frac{4}{3} - \frac{1}{3}}}{{1 + \frac{4}{3},\frac{1}{3}}} = \frac{9}{{13}}\)…
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