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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan (2 a)

Rewrite using trig identities

tan (2 a)

Rewrite as

= tan (a + a)

Use the Angle Sum identity:

tan (s + s) = \frac{tan ( s ) + tan ( s )}{1 - tan ( s ) tan ( s )}

= \frac{tan ( a ) + tan ( a )}{1 - tan ( a ) tan ( a )}

= \frac{tan ( a ) + tan ( a )}{1 - tan ( a ) tan ( a )}

Simplify

\frac{tan ( a ) + tan ( a )}{1 - tan ( a ) tan ( a )} : \frac{2 tan ( a )}{1 - tan ^{2} ( a )}

\frac{tan ( a ) + tan ( a )}{1 - tan ( a ) tan ( a )}

Add similar elements:

tan (a) + tan (a) = 2 tan (a)

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

1 - tan (a) tan (a) = 1 - tan^{2} (a)

1 - tan (a) tan (a)

tan (a) tan (a) = tan^{2} (a)

tan (a) tan (a)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (a) tan (a) = tan^{1 + 1} (a)

= tan^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= tan^{2} (a)

= 1 - tan^{2} (a)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

cos (π - θ) + sin (\frac{π}{2} + θ) = 0

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

tan (y) + cot (y) = sec (y) csc (y)

\frac{1}{sin ( x )} = csc (x)

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