Is 1+(cos^2(x))/(sin(x)-1)=-sin(x) ?

The answer to whether 1+(cos^2(x))/(sin(x)-1)=-sin(x) is True

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prove 1+(cos^2(x))/(sin(x)-1)=-sin(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

Factor

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

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

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

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

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

Cancel the common factor:

sin (x) - 1

= - (sin (x) + 1)

Negate

- (sin (x) + 1) = - sin (x) - 1

= - sin (x) - 1

= 1 + (- sin (x) - 1)

Remove parentheses:

(- a) = - a

= 1 - sin (x) - 1

Group like terms

= - sin (x) + 1 - 1

1 - 1 = 0

= - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

Is 1+(cos^2(x))/(sin(x)-1)=-sin(x) ?
The answer to whether 1+(cos^2(x))/(sin(x)-1)=-sin(x) is True