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

The answer to whether (sec^2(θ)-1)(sin^2(θ)-1)=-sin^2(θ) is True

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prove (sec^2(θ)-1)(sin^2(θ)-1)=-sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

(sec^{2} (θ) - 1) (sin^{2} (θ) - 1)

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Join

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

- 1 + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

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} ( θ ) + 1}{cos ^{2} ( θ )}

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

- 1 + sin^{2} (θ)

= - (- (- sin^{2} (θ) + 1) + 1)

= - (- (- sin^{2} (θ) + 1) + 1)

Negate

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

= (- sin^{2} (θ) + 1) - 1

Refine

= - sin^{2} (θ)

= - sin^{2} (θ)

= - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e csc (θ) = \frac{tan ( θ ) cot ( θ )}{sin ( θ )}

p ro v e cot^{2} (x) - cos^{2} (x) = cot^{2} (x) cot (x)

p ro v e cos (θ) = - \frac{3}{4}

p ro v e sin (75) = sin (\frac{150}{2})

p ro v e sec (θ) + 1 = \frac{cos ( θ ) + 1}{cos ( θ )}

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