Is tan(8θ)-tan(8θ)tan^2(4θ)=2tan(4θ) ?

The answer to whether tan(8θ)-tan(8θ)tan^2(4θ)=2tan(4θ) is True

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prove tan(8θ)-tan(8θ)tan^2(4θ)=2tan(4θ)

prove

tan (8 θ) - tan (8 θ) tan^{2} (4 θ) = 2 tan (4 θ)

True

Solution steps

tan (8 θ) - tan (8 θ) tan^{2} (4 θ) = 2 tan (4 θ)

Manipulating left side

tan (8 θ) - tan (8 θ) tan^{2} (4 θ)

Rewrite using trig identities

tan (8 θ) - tan (8 θ) tan^{2} (4 θ)

= tan (2 \cdot 4 θ) - tan (2 \cdot 4 θ) tan^{2} (4 θ)

Use the Double Angle identity:

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

= \frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} - \frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} tan^{2} (4 θ)

Simplify

\frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} - \frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} tan^{2} (4 θ) : 2 tan (4 θ)

\frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} - \frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} tan^{2} (4 θ)

\frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} tan^{2} (4 θ) = \frac{2 tan ^{3} ( 4 θ )}{1 - tan ^{2} ( 4 θ )}

\frac{2 tan ( 4 θ )}{1 - tan ^{2} ( 4 θ )} tan^{2} (4 θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 tan ( 4 θ ) tan ^{2} ( 4 θ )}{1 - tan ^{2} ( 4 θ )}

2 tan (4 θ) tan^{2} (4 θ) = 2 tan^{3} (4 θ)

2 tan (4 θ) tan^{2} (4 θ)

Apply exponent rule:

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

tan (4 θ) tan^{2} (4 θ) = tan^{1 + 2} (4 θ)

= 2 tan^{1 + 2} (4 θ)

Add the numbers:

1 + 2 = 3

= 2 tan^{3} (4 θ)

= \frac{2 tan ^{3} ( 4 θ )}{1 - tan ^{2} ( 4 θ )}

= \frac{2 tan ( 4 θ )}{- tan ^{2} ( 4 θ ) + 1} - \frac{2 tan ^{3} ( 4 θ )}{- tan ^{2} ( 4 θ ) + 1}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 tan ( 4 θ ) - 2 tan ^{3} ( 4 θ )}{1 - tan ^{2} ( 4 θ )}

Factor

2 tan (4 θ) - 2 tan^{3} (4 θ) : 2 tan (4 θ) (1 - tan^{2} (4 θ))

2 tan (4 θ) - 2 tan^{3} (4 θ)

Apply exponent rule:

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

tan^{3} (4 θ) = tan (4 θ) tan^{2} (4 θ)

= 2 tan (4 θ) - 2 tan (4 θ) tan^{2} (4 θ)

Rewrite as

= 1 \cdot 2 tan (4 θ) - 2 tan (4 θ) tan^{2} (4 θ)

Factor out common term

2 tan (4 θ)

= 2 tan (4 θ) (1 - tan^{2} (4 θ))

= \frac{2 tan ( 4 θ ) ( 1 - tan ^{2} ( 4 θ ) )}{1 - tan ^{2} ( 4 θ )}

Cancel the common factor:

1 - tan^{2} (4 θ)

= 2 tan (4 θ)

= 2 tan (4 θ)

= 2 tan (4 θ)

We showed that the two sides could take the same form

\Rightarrow True

Is tan(8θ)-tan(8θ)tan^2(4θ)=2tan(4θ) ?
The answer to whether tan(8θ)-tan(8θ)tan^2(4θ)=2tan(4θ) is True