JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The value of \({\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]\,,\,\left| x \right| < \frac{1}{2},\,x \ne 0\,,\) is equal to
- A \(\frac{\pi }{4} + \frac{1}{2}\,{\cos ^{ - 1}}\,{x^2}\)
- B \(\frac{\pi }{4} + \,{\cos ^{ - 1}}\,{x^2}\)
- C \(\frac{\pi }{4} - \frac{1}{2}\,{\cos ^{ - 1}}\,{x^2}\)
- D \(\frac{\pi }{4} - \,{\cos ^{ - 1}}\,{x^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi }{4} + \frac{1}{2}\,{\cos ^{ - 1}}\,{x^2}\)
Step-by-step Solution
Detailed explanation
Lat \({x^2} = \cos \,2\theta \,\,\,\,\,\, \Rightarrow \,\,\theta = \frac{1}{2}{\cos ^{ - 1}}{x^2}\) \( \Rightarrow {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right] = \)…
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