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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Expand

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

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

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = - sec^{2} (x), c = 1, d = - sin^{2} (x)

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

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

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

Simplify

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - sin^{2} (x) + sec^{2} (x) sin^{2} (x) - tan^{2} (x)

= - sin^{2} (x) + sec^{2} (x) sin^{2} (x) - tan^{2} (x)

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

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

Multiply fractions:

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

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

Add similar elements:

- \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} = 0

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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