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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
${d \over {dx}}\left( {{{\sec x + \tan x} \over {\sec x - \tan x}}} \right) = $
  • ✓
    ${{2\cos x} \over {{{(1 - \sin x)}^2}}}$
  • B
    ${{\cos x} \over {{{(1 - \sin x)}^2}}}$
  • C
    ${{2\cos x} \over {1 - \sin x}}$
  • D
    None of these
✓

Answer

Correct option: A.
${{2\cos x} \over {{{(1 - \sin x)}^2}}}$
a
(a) $\frac{d}{{dx}}\left( {\frac{{\sec x + \tan x}}{{\sec x - \tan x}}} \right) = \frac{d}{{dx}}\left( {\frac{{1 + \sin x}}{{1 - \sin x}}} \right)$.

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