Is cos(θ)(tan(θ)-sec(-θ))=sin(θ)-1 ?

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

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prove cos(θ)(tan(θ)-sec(-θ))=sin(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

cos (θ) (tan (θ) - sec (- θ))

Use the negative angle identity:

sec (- x) = sec (x)

= cos (θ) (- sec (θ) + tan (θ))

Express with sin, cos

(- sec (θ) + tan (θ)) cos (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Apply rule

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

= \frac{- 1 + sin ( θ )}{cos ( θ )}

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= - - 1 + sin (θ)

= - 1 + sin (θ)

= - 1 + sin (θ)

= sin (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

cos (\frac{π}{2}) = - 0

cos (x) + 1 = sin (2 x)

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

(sin (x)) (tan (x) cos (x) - cot (x) cos (x)) = 1 - 2 cos (2 x)

sin (2 0^{\circ}) = 2 cos (1 0^{\circ}) \cdot sin (1 0^{\circ})

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