Is (tan(θ))/(tan(θ)-1)= 1/(1-cot(θ)) ?

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

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prove (tan(θ))/(tan(θ)-1)= 1/(1-cot(θ))

prove

\frac{tan ( θ )}{tan ( θ ) - 1} = \frac{1}{1 - cot ( θ )}

True

Solution steps

\frac{tan ( θ )}{tan ( θ ) - 1} = \frac{1}{1 - cot ( θ )}

Manipulating left side

\frac{tan ( θ )}{tan ( θ ) - 1}

Express with sin, cos

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

Manipulating right side

\frac{1}{1 - cot ( θ )}

Express with sin, cos

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

We showed that the two sides could take the same form

\Rightarrow True

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

2 cos (x) tan (x) csc (x) = 2

sec^{2} (y) - cot^{2} (\frac{π}{2} - y) = 1

sin (\frac{π}{3} + x) - sin (\frac{π}{3} - x) = sin (x)

1 + cos (2 x) = 2 cos^{2} (x)

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