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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

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

Rewrite using trig identities

= - \frac{cos ( x )}{sin ( x )} \cdot \frac{cos ( x )}{sin ( x )}

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

- cot (x) cot (x)

Simplify

- cot (x) cot (x) : - cot^{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)

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

Add the numbers:

1 + 1 = 2

= - cot^{2} (x)

- cot^{2} (x)

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