What is the limit of the function sin x/(1-2sin^2x) -cos x/(2cos^2x-1) - sin2x/cosx, if x approaches to pi/4?

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We need to find the value of lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/2*(cos x)^2 - 1) - sin 2x / cos x.

Substituting x = pi/4, gives us an indeterminate value.

lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/(2*(cos x)^2 - 1) - sin 2x / cos x

use (1- 2*(sin x)^2) = 2*(cos x)^2 - 1 = cos 2x

=> lim x--> pi/4 [ sin x/cos 2x - cos x/cos 2x - sin 2x/cos x]

=> lim x--> pi/4 [ (sin x - cos x)/cos 2x - sin 2x/cos x]

=> lim x--> pi/4 [ (sin x - cos x)/( (cos x)^2 - (sin x)^2) - sin 2x/cos x]

=> lim x--> pi/4 [ -1/(cos x + sin x) - sin 2x/cos x]

sin pi/4 = cos pi/4 = 1/sqrt 2

substituting x = pi/4

=> -1/((1/sqrt 2) + (1/sqrt 2)) - 1/(1/sqrt 2)

=> -1/sqrt 2 - sqrt 2

=> -3/sqrt 2

The required limit is -3/sqrt 2

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