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}

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

sec (θ) cos (θ) csc (θ) = cot (θ)

cot (x) (sin (x) + tan (x)) = 1 + cos (x)

csc^{2} (x) (1 - cos^{2} (x)) = tan (42 0^{\circ})

1 - cot (x) = 1

cot (x) - csc (x) = - \frac{1}{cot ( x ) + csc ( x )}

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