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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (a)

Express with sin, cos

tan^{2} (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ^{2} ( x )}{csc ^{2} ( x )} = tan^{2} (x)

cos (4 π + x) = cos (x)

sin (x) cos (y) + sin (y) cos (x) = sin (x + y)

\frac{csc ^{2} ( x )}{sec ^{2} ( x )} = cot^{2} (x)

tan (A + B) + tan (A - B) = \frac{( 2 sin ( 2 A ) )}{( cos ( 2 A ) + cos ( 2 B ) )}

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