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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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 )}{1 - cos ( x )}

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

cos (x) - 1

= cos (x) + 1

= cos (x) + 1

= 1 - (cos (x) + 1)

- (cos (x) + 1) : - cos (x) - 1

- (cos (x) + 1)

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

= - cos (x) - 1

= 1 - cos (x) - 1

Simplify

1 - cos (x) - 1 : - cos (x)

1 - cos (x) - 1

Group like terms

= - cos (x) + 1 - 1

1 - 1 = 0

= - cos (x)

= - cos (x)

= - cos (x)

= - cos (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) csc (x) = sec (x)

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

tan (u) - sec (u) = - \frac{cos ( u )}{1 + sin ( u )}

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

(1 + tan^{2} (θ)) cos^{2} (θ) = 1

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