Is tan^2(α)(1-sin^2(α))=sin^2(α) ?

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

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prove tan^2(α)(1-sin^2(α))=sin^2(α)

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (α) (1 - sin^{2} (α))

Express with sin, cos

(1 - sin^{2} (α)) tan^{2} (α)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )} (- sin^{2} (α) + 1)

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos^{2} (α)

= sin^{2} (α)

= sin^{2} (α)

We showed that the two sides could take the same form

\Rightarrow True

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