Is (sin(θ)-1)/(cos(θ))=tan(θ)-sec(θ) ?

The answer to whether (sin(θ)-1)/(cos(θ))=tan(θ)-sec(θ) is True

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prove (sin(θ)-1)/(cos(θ))=tan(θ)-sec(θ)

prove

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

True

Solution steps

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

Manipulating right side

tan (θ) - sec (θ)

Express with sin, cos

= \frac{- 1 + sin ( θ )}{cos ( θ )}

= \frac{sin ( θ ) - 1}{cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

cot (x) \cdot sec (x) = csc (x)

2 - csc^{2} (θ) = 1 - cot^{2} (θ)

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

sin (x) = \frac{1}{csc ( x )}

cos (x) = sin (\frac{π}{2} - x)

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