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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(1 - cos^{2} (x)) cot (x)

Express with sin, cos

(1 - cos^{2} (x)) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

= sin (x) cos (x)

= sin (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

(tan (x) + cot (x))^{4} = csc^{4} (x) sec^{4} (x)

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

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

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

\frac{csc ( x ) - cot ( x )}{sec ( x ) - 1} = cot (x)

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