Is tan(θ)csc^2(θ)-tan(θ)=cot(θ) ?

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

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prove tan(θ)csc^2(θ)-tan(θ)=cot(θ)

prove

tan (θ) csc^{2} (θ) - tan (θ) = cot (θ)

True

Solution steps

tan (θ) csc^{2} (θ) - tan (θ) = cot (θ)

Manipulating left side

tan (θ) csc^{2} (θ) - tan (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1 + cot ( x )}{1 + tan ( x )} = cot (x)

cos (α + β) + cos (α - β) = 2 cos (α) cos (β)

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

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

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