STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

माना ${\cot ^{ - 1}}\frac{1}{2} = \phi  \Rightarrow \frac{1}{2} = \cot \phi  $

$ \Rightarrow \sin \phi  = \frac{1}{{\sqrt {1 + {{\cot }^2}\phi  } }} = \frac{2}{{\sqrt 5 }}$

माना ${\cos ^{ - 1}}x = \theta  \Rightarrow \sec \theta  = \frac{1}{x}$

$ \Rightarrow \tan \theta  = \sqrt {{{\sec }^2}\theta  - 1} $

$ \Rightarrow \tan \theta  = \sqrt {\frac{1}{{{x^2}}} - 1} $

$ \Rightarrow \tan \theta  = \frac{{\sqrt {1 - {x^2}} }}{x}$

अत:  $\tan \{ {\cos ^{ - 1}}(x)\}  = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right)$

$ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = \sin \left( {{{\sin }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$

$ \Rightarrow \frac{{\sqrt {1 - {x^2}} }}{x} = \frac{2}{{\sqrt 5 }} $

$\Rightarrow \sqrt {(1 - {x^2})5}  = 2x$

दोनों पक्षों का वर्ग करने पर, $x =  \pm \frac{{\sqrt 5 }}{3}$.