Is (1-cos^2(A))(1+cot^2(A))=1 ?

The answer to whether (1-cos^2(A))(1+cot^2(A))=1 is True

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prove (1-cos^2(A))(1+cot^2(A))=1

prove

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

True

Solution steps

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

Manipulating left side

(1 - cos^{2} (A)) (1 + cot^{2} (A))

Express with sin, cos

(1 + cot^{2} (A)) (1 - cos^{2} (A))

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

1 + \frac{cos ^{2} ( A )}{sin ^{2} ( A )} : \frac{sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )}

1 + \frac{cos ^{2} ( A )}{sin ^{2} ( A )}

Convert element to fraction:

1 = \frac{1 sin ^{2} ( A )}{sin ^{2} ( A )}

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

Multiply:

1 \cdot sin^{2} (A) = sin^{2} (A)

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

\frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + 1 - cos ^{2} ( A ) )}{1 - cos ^{2} ( A )}

(1 - cos^{2} (A)) (cos^{2} (A) + 1 - cos^{2} (A)) = 1 - cos^{2} (A)

(1 - cos^{2} (A)) (cos^{2} (A) + 1 - cos^{2} (A))

Simplify

cos^{2} (A) + 1 - cos^{2} (A) : 1

cos^{2} (A) + 1 - cos^{2} (A)

Group like terms

= cos^{2} (A) - cos^{2} (A) + 1

Add similar elements:

cos^{2} (A) - cos^{2} (A) = 0

= 1

= 1 \cdot (- cos^{2} (A) + 1)

Multiply:

(1 - cos^{2} (A)) \cdot 1 = (1 - cos^{2} (A))

= (1 - cos^{2} (A))

Remove parentheses:

(a) = a

= 1 - cos^{2} (A)

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)

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

20 cos^{2} (x) - cos (x) - 1 = 0

tan (\frac{y}{2}) = csc (y) - cot (y)

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

Is (1-cos^2(A))(1+cot^2(A))=1 ?
The answer to whether (1-cos^2(A))(1+cot^2(A))=1 is True