Is (sin^2(θ))/(1-sin^2(θ))=tan^2(θ) ?

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

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prove (sin^2(θ))/(1-sin^2(θ))=tan^2(θ)

prove

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

True

Solution steps

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

Manipulating right side

tan^{2} (θ)

Express with sin, cos

tan^{2} (θ)

Use the basic trigonometric identity:

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

= (\frac{sin ( θ )}{cos ( θ )})^{2}

Simplify

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan^{4} (x) = (1 - sec^{2} (x))^{2}

- 2 sin (x) = - sin (2 x)

\frac{1}{1 - \frac{1}{( 1 - \frac{1}{( 1 - c o s h ^{2} ( x ) )} )}} = cosh^{2} (x)

(\frac{sec ( θ )}{cos ( θ )}) - (\frac{tan ( θ )}{cot ( θ )}) = 1

sin (π) = sin (3 π)

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