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

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

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

prove

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

True

Solution steps

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

Manipulating right side

csc (θ) - cot (θ)

Express with sin, cos

= \frac{1 - cos ( θ )}{sin ( θ )}

Rewrite using trig identities

= \frac{sin ( θ )}{1 + cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

(cot (x) + tan (x))^{2} = csc^{2} (x) sec^{2} (x)

(sec (a)) (sin (a)) = tan (a)

cos (3 x) = 4 cos (3 x) - 3 cos (x)

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

sec^{2} (a) + csc^{2} (a) = sec^{2} (a) csc^{2} (a)

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