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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (6 x) = 2 cos^{2} (3 x) - 1

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

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

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

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

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