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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 1

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

Multiply:

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

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

Remove parentheses:

(a) = a

= 1 - cos^{2} (x)

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

tan (2 π + θ) = tan (θ)

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

cos (a) + sin (a) tan (a) = sec (a)

(csc (θ) + cot (θ)) (1 - cos (θ)) = sin (θ)

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