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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

= cos (x) sin (x)

Use the Double Angle identity:

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

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

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

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

Manipulating right side

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

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

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

(1 - tan (x)) (1 - cot (x)) = 2 - sec (x) csc (x)

(1 - sin^{2} (θ)) (1 + tan^{2} (θ)) = 1

sin (x) + cos (x) \cdot cot (x) = csc (x)

cos (4 x) = 1 - 8 sin^{2} (x) cos^{2} (x)

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