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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (x) - csc (x) cot (x)

Express with sin, cos

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

Factor

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

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

Rewrite using trig identities

= - \frac{cos ^{3} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1 + tan ( x )}{sin ( x )} - sec (x) = csc (x)

\frac{csc ( β )}{cot ( β )} = sec (β)

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

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

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