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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

cos (x) + cos (x) cot^{2} (x) : cos (x) (1 + cot^{2} (x))

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

Express with sin, cos

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

Rewrite using trig identities

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

Manipulating right side

cot (x) csc (x)

Express with sin, cos

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

We showed that the two sides could take the same form

\Rightarrow True

tan (a) = tan (a) csc^{2} (a) + cot (- a)

7 csc^{2} (x) - 5 cot^{2} (x) = 2 csc^{2} (x) + 5

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

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

tan (θ) (cot (θ) + tan (θ)) = sec^{2} (θ)

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