Is (sin^2(θ))/(1-cos(θ))=1+cos(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (sin^2(θ))/(1-cos(θ))=1+cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (θ)

Factor out common term

- 1

= - (cos^{2} (θ) - 1)

Factor

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

cos^{2} (θ) - 1

Rewrite

1

1^{2}

= cos^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

Cancel

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

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

- cos (θ) + 1 = - (cos (θ) - 1)

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

Refine

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

Cancel the common factor:

cos (θ) - 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

csc (2 θ) = \frac{1}{2} sec (θ) csc (θ)

sin (x) (1 + cot (x)) = sin (x) + cos (x)

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

cot (\frac{π}{2} - x) csc (x) = sec (x)

csc^{2} (x) + sec^{2} (x) = csc^{2} (x) sec^{2} (x)

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