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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(tan (x) - 1) cos (x)

Express with sin, cos

(- 1 + tan (x)) cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= - - cos (x) + sin (x)

= - cos (x) + sin (x)

= - cos (x) + sin (x)

= sin (x) - cos (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (θ - \frac{π}{2}) = - cos (θ)

sec (x) - cos (x) = tan (x) \cdot sin (x)

\frac{cot ( x ) - 1}{1 - tan ( x )} = cot (x)

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

2 sin (2 x) sin (2 x - \frac{π}{2}) = - cos (4 x - \frac{π}{2})

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