MCQ
If $y = {\log _{\cos x}}\sin x$, then ${{dy} \over {dx}}$ is equal to
  • ${{\cot x\log \cos x + \tan x\log \sin x} \over {{{(\log \cos x)}^2}}}$
  • B
    ${{\tan x\log \cos x + \cot x\log \sin x} \over {{{(\log \cos x)}^2}}}$
  • C
    ${{\cot x\log \cos x + \tan x\log \sin x} \over {{{(\log \sin x)}^2}}}$
  • D
    None of these

Answer

Correct option: A.
${{\cot x\log \cos x + \tan x\log \sin x} \over {{{(\log \cos x)}^2}}}$
a
(a) We have $y = {\log _{\cos x}}\sin x = \frac{{\log \sin x}}{{\log \cos x}}$

$\therefore \frac{{dy}}{{dx}} = \frac{{\cot x.\log \cos x + (\log \sin x)\tan x}}{{{{(\log \cos x)}^2}}}$.

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