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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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