Is cot^2(θ)-cos^2(θ)=cos^2(θ)cot^2(θ) ?

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

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prove cot^2(θ)-cos^2(θ)=cos^2(θ)cot^2(θ)

prove

cot^{2} (θ) - cos^{2} (θ) = cos^{2} (θ) cot^{2} (θ)

True

Solution steps

cot^{2} (θ) - cos^{2} (θ) = cos^{2} (θ) cot^{2} (θ)

Manipulating left side

cot^{2} (θ) - cos^{2} (θ)

Express with sin, cos

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

Factor

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

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

Rewrite using trig identities

= \frac{cos ^{4} ( θ )}{sin ^{2} ( θ )}

Rewrite using trig identities

cos^{2} (θ) cot^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( A )}{sec ( A )} = sin (A)

tan (8 x) - tan (8 x) tan^{2} (4 x) = 2 tan (4 x)

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

cot (2 u) = \frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

\frac{sin ( θ ) csc ( θ )}{cot ( θ )} = tan (θ)

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