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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

1 - (1 - 2 sin^{2} (x)) : 2 sin^{2} (x)

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (x)

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

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

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

(cos (2 x) - sin (2 x))^{2} = 1 - sin (4 x)

sec^{2} (x) csc^{2} (x) = csc^{2} (x) + sec^{2} (x)

cot (π - θ) = - cot (θ)

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

\frac{( sec ( x ) - 1 )}{1 - cos ( x )} = sec (x)

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