Is (csc(θ)-cot(θ))/(sec(θ)-1)=cot(θ) ?

The answer to whether (csc(θ)-cot(θ))/(sec(θ)-1)=cot(θ) is True

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prove (csc(θ)-cot(θ))/(sec(θ)-1)=cot(θ)

prove

\frac{csc ( θ ) - cot ( θ )}{sec ( θ ) - 1} = cot (θ)

True

Solution steps

\frac{csc ( θ ) - cot ( θ )}{sec ( θ ) - 1} = cot (θ)

Manipulating left side

\frac{csc ( θ ) - cot ( θ )}{sec ( θ ) - 1}

Express with sin, cos

= \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

sin^{2} (z) + cos^{2} (z) = 1

tan (x) sin (x) = cos (x)

cot (θ) = \frac{1}{tan ( θ )}

cos^{2} (θ) tan^{2} (θ) = sin^{2} (θ)

1 - cot (x) = csc^{2} (x) - 2 cot (x)

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