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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan (2 x)

Rewrite using trig identities

tan (2 x)

Rewrite as

= tan (x + x)

Use the Angle Sum identity:

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

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

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

Simplify

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

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

Add similar elements:

tan (x) + tan (x) = 2 tan (x)

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

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

1 - tan (x) tan (x)

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

tan (x) tan (x)

Apply exponent rule:

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

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

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

= 1 - tan^{2} (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (x) = \frac{1 + cos ( 2 x )}{2}

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

cot^{2} (y) (sec^{2} (y) - 1) = 1

tan (x - \frac{π}{2}) = - cot (x)

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

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