1.

\( \frac{\cos 2 x}{1+\sin 2 x}=\tan (\pi / 4-x) \)

Answer» LHS

=Cos2x/(1+sin2x)

=(Cos^2 x -sin^2 x)/(sin^2 x+ cos^2 x+2sin x cos x)

Here,

cos 2x= cos^2 x- sin^2 x

1=sin^2 x+cos^2 x

sin 2x=2sin x cos x

So,

=(Cos x-sin x)(cos x+sin x)/(cos x+sin x)^2

Because we know that

A^2-B^2=(A-B)(A+B)

A^2+2AB+B^2=(A+B)^2

Now,

Common terms cancel out

=(Cos x-sin x)/(cos x+ sin x)

Dividing by  cos x in numerator and denominator

We get

=(1-sin x/cos x)/(1+sin x/cos x)

=(1-tan x)/(1+tan x)

RHS

tan(π/4-x)

Using the following formula

tan(A-B)={tanA - tanB)/{1+tanAtanB)

tan(π/4-x)={tan π/4 -tanx}/{1+tan π/4.tanx}

tan(π/4-x)={1-tanx}/{1+tanx}

Here, tan π/4=1

Hence proved


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