Is 1/(sin(θ))-sin(θ)=cot(θ)cos(θ) ?

The answer to whether 1/(sin(θ))-sin(θ)=cot(θ)cos(θ) is True

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prove 1/(sin(θ))-sin(θ)=cot(θ)cos(θ)

prove

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

True

Solution steps

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

Manipulating right side

cot (θ) cos (θ)

Express with sin, cos

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

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

Expand

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

= \frac{1}{sin ( θ )} - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (θ) cos (θ) (tan (θ) + cot (θ)) = 1

tan (x) = \frac{sec ( x )}{csc ( x )}

\frac{csc ( θ )}{cot ( θ )} = sec (θ)

\frac{sin ( t )}{1 - cos ( t )} = csc (t) + cot (t)

coth^{2} (x) - csch^{2} (x) = 1

Is 1/(sin(θ))-sin(θ)=cot(θ)cos(θ) ?
The answer to whether 1/(sin(θ))-sin(θ)=cot(θ)cos(θ) is True