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

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

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (x) - 1

Express with sin, cos

- 1 + sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}

= - 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 )}

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

Rewrite using trig identities

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= sin^{2} (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) = - \frac{3}{4}

sin (75) = sin (\frac{150}{2})

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

cos (4 x) cos (x) - sin (4 x) - sin (x) = 0

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

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