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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan (2 u)

Rewrite using trig identities

tan (2 u)

Rewrite as

= tan (u + u)

Use the Angle Sum identity:

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

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

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

Simplify

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

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

Add similar elements:

tan (u) + tan (u) = 2 tan (u)

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

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

1 - tan (u) tan (u)

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

tan (u) tan (u)

Apply exponent rule:

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

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

= tan^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= tan^{2} (u)

= 1 - tan^{2} (u)

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{3 sec ( θ ) - 3}{1 - cos ( θ )} = 3 sec (θ)

cot (\frac{π}{2} - θ) = tan (θ)

cos (\frac{π}{2} - a) = sin (a)

\frac{sin ( x + y )}{sin ( x ) cos ( y )} = 1 + cot (x) tan (y)

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

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