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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( y ) + 1}{cos ^{2} ( y )}

Multiply:

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

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

- 1 + sin^{2} (y)

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

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

Negate

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

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

Refine

= - sin^{2} (y)

= - sin^{2} (y)

= - sin^{2} (y)

We showed that the two sides could take the same form

\Rightarrow True

(1 + tan (β)) (1 - tan (β)) = 2 - sec^{2} (β)

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

cos^{2} (x) - cos^{2} (9 0^{\circ} - x) = 1 - 2 sin (x)

sec^{2} (t) csc^{2} (t) = sec^{2} (t) + csc^{2} (t)

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

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