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

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

prove

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

True

Solution steps

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

Manipulating right side

tan^{2} (x)

Express with sin, cos

tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

cos^{2} (x) = 1 - 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

sin (- 3 0^{\circ}) = - sin (3 0^{\circ})

\frac{1}{1 - cot ( x )} = 1 - cot (x)

cos (\frac{1}{2} π - x) = sin (x)

sin (x) - cos (x) + 1 = cos (x)

cos (12 x) = 1 - 2 sin^{2} (6 x)

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