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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

cot^{2} (θ) - cos^{2} (θ)

Express with sin, cos

- cos^{2} (θ) + cot^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

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

Factor

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

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

Factor

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

cos^{2} (θ) - cos^{2} (θ) sin^{2} (θ)

Factor out common term

cos^{2} (θ)

= cos^{2} (θ) (1 - sin^{2} (θ))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

cos^{2} (θ) cos^{2} (θ) = cos^{4} (θ)

cos^{2} (θ) cos^{2} (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{2 + 2} (θ)

Add the numbers:

2 + 2 = 4

= cos^{4} (θ)

= \frac{cos ^{4} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{4} ( θ )}{sin ^{2} ( θ )}

Rewrite using trig identities

= \frac{cos ^{3} ( θ )}{sin ( θ )} \cdot \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

\frac{cos ^{3} ( θ ) cot ( θ )}{sin ( θ )}

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

Use the basic trigonometric identity:

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

cos^{2} (θ) cot (θ) cot (θ)

Simplify

cos^{2} (θ) cot (θ) cot (θ) : cos^{2} (θ) cot^{2} (θ)

cos^{2} (θ) cot (θ) cot (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (θ) cot (θ) = cot^{1 + 1} (θ)

= cos^{2} (θ) cot^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ) cot^{2} (θ)

cos^{2} (θ) cot^{2} (θ)

cos^{2} (θ) cot^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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