Is (2tan(θ))/(1-tan^2(θ))=tan(2θ) ?

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

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prove (2tan(θ))/(1-tan^2(θ))=tan(2θ)

prove

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

True

Solution steps

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

Manipulating right 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

cos (x) sin (x) = \frac{1}{2} sin (2 x)

\frac{1 + tan ( x )}{1 - tan ( x )} = tan (x + \frac{π}{4})

4 sin (x) cos (x) = 2 sin (2 x)

sin^{2} (a) cos^{2} (a) + cos^{4} (a) = cos^{2} (a)

\frac{cot ( θ )}{cos ( θ )} = csc (θ)

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