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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

\frac{- 1 + sec ( x )}{1 - cos ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

- 1 + \frac{1}{cos ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

- cos (x) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) tan (x) + cos (x) = sec (x)

csc (θ) cos (θ) = cot (θ)

csc^{2} (x) tan^{2} (x) - 1 = tan^{2} (x)

sin (x) - sin (x) cos^{2} (x) = sin^{3} (x)

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

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