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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ( θ ) - 1}{1 - cos ( θ )}

Express with sin, cos

\frac{- 1 + sec ( θ )}{1 - cos ( θ )}

Use the basic trigonometric identity:

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

= \frac{- 1 + \frac{1}{c o s ( θ )}}{1 - cos ( θ )}

Simplify

\frac{- 1 + \frac{1}{c o s ( θ )}}{1 - cos ( θ )} : \frac{1}{cos ( θ )}

\frac{- 1 + \frac{1}{c o s ( θ )}}{1 - cos ( θ )}

Join

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

- 1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{- cos ( θ ) + 1}{cos ( θ )}

= \frac{\frac{- c o s ( θ ) + 1}{c o s ( θ )}}{1 - cos ( θ )}

Apply the fraction rule:

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

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

Cancel the common factor:

- cos (θ) + 1

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{cot ( θ ) tan ( θ )}{sec ( θ )} = cos (θ)

sin (a) cos (a) = cos^{2} (a) tan (a)

(cos (x)) (tan (x)) (csc (x)) = 1

\frac{csc ( x ) tan ( x )}{1 + tan ( x )} = cos (x)

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