MCQ
જો $y = {\tan ^{ - 1}}(\sec x - \tan x)$ તો ${{dy} \over {dx}} = $
- A$2$
- ✓$ -0.5$
- C$\frac{1}{2}$
- D$-2$
$\frac{{dy}}{{dx}} = \frac{1}{{1 + {{(\sec x - \tan x)}^2}}}(\sec x\tan x - {\sec ^2}x)$
$\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}x.{{\sec }^2}x(\sin x - 1)}}{{{{(1 - \sin x)}^2} + {{\cos }^2}x}}$
$\frac{{dy}}{{dx}} = \frac{{\sin x - 1}}{{1 - 2\sin x + {{\sin }^2}x + {{\cos }^2}x}} $
$= \frac{{\sin x - 1}}{{2(1 - \sin x)}} = - \frac{1}{2}.$
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