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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $y = {\tan ^{ - 1}}(\sec x - \tan x)$ then ${{dy} \over {dx}} = $
  • A
    $2$
  • ✓
    $ -0.5$
  • C
    $\frac{1}{2}$
  • D
    $-2$
✓

Answer

Correct option: B.
$ -0.5$
b
(b) $y = {\tan ^{ - 1}}(\sec x - \tan x)$

$\frac{{dy}}{{dx}} = \frac{1}{{1 + {{(\sec x - \tan x)}^2}}}(\sec x\tan x - {\sec ^2}x)$

$\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}x.{{\sec }^2}x(\sin x - 1)}}{{{{(1 - \sin x)}^2} + {{\cos }^2}x}}$

$\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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