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

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

\frac{1}{- sin ( x )} = sec (x) tan (x)

6 cos (θ) = 6 cos (- θ)

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

cos (x + (\frac{3 π}{2})) = sin (x)

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