Is csc(θ)-(sin(θ))/(1+cos(θ))=cot(θ) ?

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

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prove csc(θ)-(sin(θ))/(1+cos(θ))=cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

csc (θ) - \frac{sin ( θ )}{1 + cos ( θ )}

Express with sin, cos

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

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

sin^{2} (a) + sin^{2} (a) tan^{2} (a) = tan^{2} (a)

sin (b) + cos (b) cot (b) = csc (b)

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

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