Is (cot(x)-1)/(cot(x)+1)=(cos(2x))/(1+sin(2x)) ?

The answer to whether (cot(x)-1)/(cot(x)+1)=(cos(2x))/(1+sin(2x)) is True

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prove (cot(x)-1)/(cot(x)+1)=(cos(2x))/(1+sin(2x))

prove

\frac{cot ( x ) - 1}{cot ( x ) + 1} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

True

Solution steps

\frac{cot ( x ) - 1}{cot ( x ) + 1} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

Manipulating left side

\frac{cot ( x ) - 1}{cot ( x ) + 1}

Express with sin, cos

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

Multiply by

\frac{cos ( 2 x ) ( cos ( x ) - sin ( x ) )}{cos ( 2 x ) ( cos ( x ) - sin ( x ) )}

= \frac{( cos ( x ) - sin ( x ) ) ( cos ( x ) - sin ( x ) ) cos ( 2 x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) ) cos ( 2 x )}

Expand

(cos (x) - sin (x)) (cos (x) - sin (x)) cos (2 x) : cos^{2} (x) cos (2 x) - 2 cos (2 x) cos (x) sin (x) + sin^{2} (x) cos (2 x)

= \frac{cos ( 2 x ) cos ^{2} ( x ) + cos ( 2 x ) sin ^{2} ( x ) - 2 cos ( 2 x ) cos ( x ) sin ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) ) cos ( 2 x )}

Rewrite using trig identities

= \frac{cos ( 2 x )}{1 + sin ( 2 x )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )} = \frac{1}{cos ^{2} ( u ) - sin ^{2} ( u )}

csc (θ) + cot (θ) = \frac{sin ( θ )}{1 - cos ( θ )}

1 + tan^{2} (- x) = sec^{2} (x)

tan (\frac{π}{2} - x) = cot (x)

sec (θ) = \frac{tan ( θ )}{sin ( θ )}

Is (cot(x)-1)/(cot(x)+1)=(cos(2x))/(1+sin(2x)) ?
The answer to whether (cot(x)-1)/(cot(x)+1)=(cos(2x))/(1+sin(2x)) is True