Is cos(x)(tan(x)-sec(-x))=sin(x)-1 ?

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

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prove cos(x)(tan(x)-sec(-x))=sin(x)-1

prove

cos (x) (tan (x) - sec (- x)) = sin (x) - 1

True

Solution steps

cos (x) (tan (x) - sec (- x)) = sin (x) - 1

Manipulating left side

cos (x) (tan (x) - sec (- x))

Use the negative angle identity:

sec (- x) = sec (x)

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

Express with sin, cos

(- sec (x) + tan (x)) cos (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

Simplify

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

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

Apply rule

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= - - 1 + sin (x)

= - 1 + sin (x)

= - 1 + sin (x)

= sin (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

cot (u) - csc (u) = - \frac{sin ( u )}{1 + cos ( u )}

sec^{2} (θ) - tan^{2} (θ) = 1

\frac{1 + tan ^{2} ( x )}{tan ^{2} ( x )} = csc^{2} (x)

\frac{1}{sec ^{2} ( x )} + \frac{1}{csc ^{2} ( x )} = 1

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