JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The value of \({\sin ^{ - 1}}\,\left( {\frac{{12}}{{13}}} \right) - {\sin ^{ - 1}}\,\left( {\frac{3}{5}} \right)\) is equal to
- A \(\pi - {\cos ^{ - 1}}\,\left( {\frac{{33}}{{65}}} \right)\)
- B \(\pi - {\sin ^{ - 1}}\,\left( {\frac{{63}}{{65}}} \right)\)
- C \(\frac{\pi }{2} - {\cos ^{ - 1}}\,\left( {\frac{9}{{65}}} \right)\)
- D \(\frac{\pi }{2} - {\sin ^{ - 1}}\,\left( {\frac{56}{{65}}} \right)\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi }{2} - {\sin ^{ - 1}}\,\left( {\frac{56}{{65}}} \right)\)
Step-by-step Solution
Detailed explanation
\({\sin ^{ - 1}}\left( {\frac{{12}}{{13}}} \right) - {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\) \({\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right)\)…
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