JEE Mains · Maths · STD 12 - 6. Application of derivatives
Consider :\(f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right),x \in \left( {0,\frac{\pi }{2}} \right)\) A normal to \(y = f\left( x \right)\) at \(x = \frac{\pi }{6}\) also passes through the point :
- A \(\;\left( {\frac{\pi }{6},0} \right)\)
- B \(\left( {\frac{\pi }{4},0} \right)\)
- C \(\left( {0,0} \right)\)
- D \(\;\left( {0,\frac{{2\pi }}{3}} \right)\)
Answer & Solution
Correct Answer
(D) \(\;\left( {0,\frac{{2\pi }}{3}} \right)\)
Step-by-step Solution
Detailed explanation
\(f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 + \sin x}}} } \right)\) where \(x \in \left( {0,\frac{\pi }{2}} \right)\) \( = {\tan ^{ - 1}}\left( {\sqrt {\frac{{{{\left( {1 + \sin x} \right)}^2}}}{{1 + \sin x}}} } \right)\)…
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