Is ((sec(θ)-1))/((1-cos(θ)))=sec(θ) ?

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

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

prove

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

True

Solution steps

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

sec (θ)

sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (0) tan (0) = sin (0)

cos (\frac{π}{6}) = \frac{4}{3} cos^{3} (\frac{π}{6})

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

4 sin^{2} (x) cos^{2} (x) = sin^{2} (2 x)

tan (- \frac{π}{4}) = - 1

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