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JEE Mains · Maths · STD 12 - 6. Application of derivatives

\(f(x)=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), x \in\left(0, \frac{\pi}{2}\right)\) पर विचार कीजिए।
\(y=f(x)\) के बिंदु \(x=\frac{\pi}{6}\) पर खींचा गया अभिलंब निम्न बिंदु से भी होकर जाता हैः

  1. A \(\;\left( {\frac{\pi }{6},0} \right)\)
  2. B \(\left( {\frac{\pi }{4},0} \right)\)
  3. C \(\left( {0,0} \right)\)
  4. D \(\;\left( {0,\frac{{2\pi }}{3}} \right)\)
Verified Solution

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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