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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

csc^{2} (x) (sin^{2} (x) - 1)

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(- a) = - a

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

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

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

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

Rewrite using trig identities

\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

cos (3 x) = cos (x) (4 cos^{2} (x) - 3)

\frac{1}{sin ( x ) cot ( x )} = sec (x)

csc (θ) + sin (θ) = \frac{2 - cos ^{2} ( θ )}{sin ( θ )}

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

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

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