2. inverse trigonometric function — ગણિત ધોરણ 12 સાયન્સ — Question

Now, $\cos ec \theta =\sqrt {1+ {\cot ^2 \theta}} = \sqrt {{x^2} +1} $.

$\therefore \,\,\,\sin \theta  = \frac{1}{{\cos ec\,\theta }} = \frac{1}{{\sqrt {1 + {x^2}} }}\,\, \Rightarrow \,\theta  = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$

Hence $\sin \,({\cot ^{ - 1}}x)\, = \sin \,\left( {{{\sin }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right)$

$ = \frac{1}{{\sqrt {1 + {x^2}} }} = {(1 + {x^2})^{ - 1/2}}$