Is cot(x)-1=csc(x)(cos(x)-sin(x)) ?

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

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prove cot(x)-1=csc(x)(cos(x)-sin(x))

prove

cot (x) - 1 = csc (x) (cos (x) - sin (x))

True

Solution steps

cot (x) - 1 = csc (x) (cos (x) - sin (x))

Manipulating left side

cot (x) - 1

Express with sin, cos

- 1 + cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= (cos (x) - sin (x)) csc (x)

= csc (x) (cos (x) - sin (x))

We showed that the two sides could take the same form

\Rightarrow True

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