Is cos(θ)cot(θ)=csc(θ)-sin(θ) ?

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

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prove cos(θ)cot(θ)=csc(θ)-sin(θ)

prove

cos (θ) cot (θ) = csc (θ) - sin (θ)

True

Solution steps

cos (θ) cot (θ) = csc (θ) - sin (θ)

Manipulating right side

csc (θ) - sin (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

cos (θ) cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

1 - tanh^{2} (x) = sech^{2} (x)

sin (x) (tan (x) + cot (x)) = sec (x)

cos (π - x) = - cos (x)

\frac{1 + sin ( θ )}{1 - sin ( θ )} - \frac{1 - sin ( θ )}{1 + sin ( θ )} = 4 tan (θ) sec (θ)

cos (7 x) + cos (3 x) = 2 cos (5 x) cos (2 x)

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