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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

= \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 )}

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

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

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

Factor

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

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

Factor

cos^{4} (a) + cos^{2} (a) sin^{2} (a) : cos^{2} (a) (cos^{2} (a) + sin^{2} (a))

cos^{4} (a) + cos^{2} (a) sin^{2} (a)

Apply exponent rule:

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

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

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

Factor out common term

cos^{2} (a)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

cot (a) cot (a)

Simplify

cot (a) cot (a) : cot^{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)

= cot^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cot^{2} (a)

cot^{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)+cos^2(a)cot^2(a)=cot^2(a) ?
The answer to whether cos^2(a)+cos^2(a)cot^2(a)=cot^2(a) is True