Is cos^2(x)csc(x)-csc(x)=-sin(x) ?

The answer to whether cos^2(x)csc(x)-csc(x)=-sin(x) is True

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prove cos^2(x)csc(x)-csc(x)=-sin(x)

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (x) csc (x) - csc (x)

Express with sin, cos

- csc (x) + cos^{2} (x) csc (x)

Use the basic trigonometric identity:

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

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

Simplify

- \frac{1}{sin ( x )} + cos^{2} (x) \frac{1}{sin ( x )} : \frac{- 1 + cos ^{2} ( x )}{sin ( x )}

- \frac{1}{sin ( x )} + cos^{2} (x) \frac{1}{sin ( x )}

cos^{2} (x) \frac{1}{sin ( x )} = \frac{cos ^{2} ( x )}{sin ( x )}

cos^{2} (x) \frac{1}{sin ( x )}

Multiply fractions:

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

= \frac{1 \cdot cos ^{2} ( x )}{sin ( x )}

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( x )}{sin ( x )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{- 1 + cos ^{2} ( x )}{sin ( x )}

Use the Pythagorean identity:

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

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

= \frac{- sin ^{2} ( x )}{sin ( x )}

Simplify

\frac{- sin ^{2} ( x )}{sin ( x )} : - sin (x)

\frac{- sin ^{2} ( x )}{sin ( x )}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{sin ^{2} ( x )}{sin ( x )}

Cancel the common factor:

sin (x)

= - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

Is cos^2(x)csc(x)-csc(x)=-sin(x) ?
The answer to whether cos^2(x)csc(x)-csc(x)=-sin(x) is True