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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

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

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

Factor

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

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

Factor out common term

cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

cos^{2} (x) cos^{2} (x)

Apply exponent rule:

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

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

= cos^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= cos^{4} (x)

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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