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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

- 2 tan^{2} (0) = 2 tan (0)

2 cot (x) sin (x) = 2 cos (x)

csc^{2} (x) (sin^{2} (x) - 1) = - cot^{2} (x)

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