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Question

Mathematics Question on Trigonometric Functions

The value of loge2ddx(logcosxcosecx)\log_e2\frac{d}{dx}(\log_{cos⁡ x}\cosec x) at x=π4x=\frac{\pi}{4} is

A

22-2\sqrt2

B

222\sqrt2

C

4-4

D

44

Answer

44

Explanation

Solution

The correct answer is (D) : 4
Let
f(x)=logcosxcosecxf(x) = \log_{cos x}\cosec x
=logcosecx/logcosx= \frac{\log \cosec x/}{\log \cos x}
f(x)=logcosx.sinx.(cosecxcotxlogcosecx.1cosxsinx)(logcosx)2f'(x) = \frac{\log \cos x.\sin x.(-\cosec x \cot x-\log \cosec x . \frac{1}{\cos x}-\sin x)}{(\log\cos x)^2}
at x =π4= \frac{π}{4}
f(π4)=log(12)+log2(log12)2f'(\frac{\pi}{4}) = \frac{-log(\frac{1}{\sqrt{2}})+log\sqrt2}{(log\frac{1}{\sqrt2})^2}
=2log2= \frac{2}{log\sqrt{2}}
loge2f(x) at x=π4=4∴ \log_e 2f' (x)\ at\ x = \frac{\pi}{4} = 4