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

- sec (θ) + tan (θ)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - \frac{1}{cos ( θ )} + tan (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + sin ( θ )}{cos ( θ )}

= \frac{- 1 + sin ( θ )}{cos ( θ )}

= \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 )}

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

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