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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Use the negative angle identity:

sec (- x) = sec (x)

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

1 - cos (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (x) + tan (x) = (sec (x)) csc (x)

1 - cot^{2} (θ) = 2 - csc^{2} (θ)

cos^{4} (x) - sin^{4} (x) - 1 = - 2 sin^{2} (x)

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

(sin (θ))^{2} + (cos (θ))^{2} = 1

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