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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (x)

Factor out common term

- 1

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Cancel the common factor:

cos (x) - 1

= - (cos (x) + 1)

Negate

- (cos (x) + 1) = - cos (x) - 1

= - cos (x) - 1

= - cos (x) - 1

= - cos (x) - 1

= - 1 - cos (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{7 cot ( x )}{sec ( x )} = 7 csc (x) - 7 sin (x)

sin (θ + \frac{π}{3}) = sin (θ) + sin (\frac{π}{3})

(4 sin (θ) - 4 cos (θ))^{2} = 16 - 16 sin (2 θ)

arcsec (x) = cos (x)

(csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

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