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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

= cos (x) sin (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (x) = \frac{1}{tan ( x )}

(1 - cos (θ)) (1 + cos (θ)) = \frac{1}{csc ^{2} ( θ )}

sec (- x) sin (x) = tan (x)

cos (θ) csc (θ) tan (θ) = 1

cot^{2} (a) = cos^{2} (a) + (cot (a) cos (a))^{2}

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