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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Multiply by

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(1 - sin (x)) (1 + sin (x)) = 1 - sin^{2} (x)

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

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

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

Cancel the common factor:

cos (2 x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

sec (2 π) = 1 - tan^{2} (2 π)

tan^{2} (θ) cos^{2} (θ) = 1 - cos^{2} (θ)

\frac{sin ( x )}{sin ( 2 x )} = \frac{sec ( x )}{2}

sin (2 x) = 2 sin (x) \cdot cos (x)

tan (a) \frac{1}{sin ^{2} ( a )} = csc (a) sec (a)

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