Is tan(2θ)=(2tan(θ))/(1-tan^2(θ)) ?

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

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prove tan(2θ)=(2tan(θ))/(1-tan^2(θ))

prove

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

True

Solution steps

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

Manipulating left side

tan (2 θ)

Rewrite using trig identities

tan (2 θ)

Rewrite as

= tan (θ + θ)

Use the Angle Sum identity:

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

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

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

Simplify

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

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

Add similar elements:

tan (θ) + tan (θ) = 2 tan (θ)

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

1 - tan (θ) tan (θ) = 1 - tan^{2} (θ)

1 - tan (θ) tan (θ)

tan (θ) tan (θ) = tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

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

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

= 1 - tan^{2} (θ)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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