Is (sin^2(x)-1)(sec^2(x)-1)=-sin^2(x) ?

The answer to whether (sin^2(x)-1)(sec^2(x)-1)=-sin^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 (sin^2(x)-1)(sec^2(x)-1)=-sin^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

- 1 + sin^{2} (x)

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

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

Negate

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

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

Refine

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) = sec^{2} (x) - 1

\frac{1}{cot ( x ) + 1} + \frac{1}{tan ( x ) + 1} = 1

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

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

csc (x) tan (x) = sec (x)

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