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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

cot^{2} (a) - cos^{2} (a)

Express with sin, cos

- cos^{2} (a) + cot^{2} (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

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

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

Factor

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

cos^{2} (a) - cos^{2} (a) sin^{2} (a)

Factor out common term

cos^{2} (a)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

cos^{2} (a) cos^{2} (a) = cos^{4} (a)

cos^{2} (a) cos^{2} (a)

Apply exponent rule:

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

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

= cos^{2 + 2} (a)

Add the numbers:

2 + 2 = 4

= cos^{4} (a)

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

cos^{2} (a) cot (a) cot (a)

Simplify

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

cos^{2} (a) cot (a) cot (a)

Apply exponent rule:

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

cot (a) cot (a) = cot^{1 + 1} (a)

= cos^{2} (a) cot^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cos^{2} (a) cot^{2} (a)

cos^{2} (a) cot^{2} (a)

cos^{2} (a) cot^{2} (a)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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