Is (1-cos(θ))(1+cos(θ))(1+cot^2(θ))=1 ?

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

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prove (1-cos(θ))(1+cos(θ))(1+cot^2(θ))=1

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

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

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

Simplify

cos^{2} (θ) + 1 - cos^{2} (θ) : 1

cos^{2} (θ) + 1 - cos^{2} (θ)

Group like terms

= cos^{2} (θ) - cos^{2} (θ) + 1

Add similar elements:

cos^{2} (θ) - cos^{2} (θ) = 0

= 1

= 1 \cdot (cos (θ) + 1) (- cos (θ) + 1)

Multiply:

(1 - cos (θ)) \cdot 1 = (1 - cos (θ))

= (cos (θ) + 1) (- cos (θ) + 1)

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

Factor

1 - cos (θ) : - (cos (θ) - 1)

1 - cos (θ)

Factor out common term

- 1

= - (cos (θ) - 1)

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

Cancel the common factor:

(1 + cos (θ)) (cos (θ) - 1)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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