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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - sec (x)

Express with sin, cos

1 - sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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{cos ( x ) - 1}{cos ( x )}

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e (1 + sec (θ)) (1 - cos (θ)) = sec (θ) - cos (θ)

p ro v e 6 cos^{2} (x) - 6 sin^{2} (x) = 6 - 12 sin^{2} (x)

p ro v e cos^{8} (x) = sin^{5} (x) cos^{3} (x)

p ro v e 2 sec (2 x) = tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x)

p ro v e sin (x) - csc (x) = - (cos (x)) (cot (x))

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