Is tan^2(α)=(sin^2(α))/(1-sin^2(α)) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (α)

Express with sin, cos

tan^{2} (α)

Use the basic trigonometric identity:

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

= (\frac{sin ( α )}{cos ( α )})^{2}

Simplify

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

(\frac{sin ( α )}{cos ( α )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( α )}{cos ^{2} ( α )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

tan (x) sec (x) (csc (x) - sin (x)) = 1

csc (x) cos (x) cot (x) = sin (x)

2 cos (x) + 2 sin (x) tan (x) = 2 sec (x)

sin^{4} (x) + cos^{4} (x) = 1

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