Prove that tan4x=4tanx(1-tan²x)/1-6tan²x+tan⁴x.
I think the best way to start would be
tan 4x = 2*tan 2x/(1-tan^2 2x)
= 2(2tanx/(1-tan^2x)) / (1-(2tanx/(1-tan^2x))^2)
Now place top and bottom over a common denominator of (1-tan^2x)^2
and you should see things going where you want them.
To prove that $\tan 4x = \frac{4\tan x (1- \tan^2 x)}{1-6\tan^2 x + \tan^4 x}$, we can use the fundamental trigonometric identity $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Step 1: Start by converting the expression on the right-hand side to the sum of angles form.
$\tan 4x = \frac{4\tan x (1- \tan^2 x)}{1-6\tan^2 x + \tan^4 x}$
Let's simplify the numerator and denominator separately:
Numerator:
$4\tan x (1- \tan^2 x) = 4 \tan x - 4 \tan^3 x$
Denominator:
$1-6\tan^2 x + \tan^4 x = \tan^4 x - 6 \tan^2 x + 1$
Step 2: Replace the numerator and denominator in the original expression with their respective simplified forms.
$\tan 4x = \frac{4 \tan x - 4 \tan^3 x}{\tan^4 x - 6 \tan^2 x + 1}$
Step 3: Apply the sum of angles formula.
Let's set $A = x$ and $B = 3x$.
Using the sum of angles formula, we have:
$\tan (4x) = \tan (x + 3x) = \frac{\tan x + \tan (3x)}{1 - \tan x \tan (3x)}$
Now, we need to express $\tan (3x)$ in terms of $\tan x$.
Step 4: Expand $\tan (3x)$ using the triple-angle formula.
$\tan (3x) = \tan (x + 2x) = \frac{\tan x + \tan (2x)}{1 - \tan x \tan (2x)}$
Step 5: Use the double-angle formula to express $\tan (2x)$ in terms of $\tan x$.
$\tan (2x) = \frac{2 \tan x}{1 - \tan^2 x}$
Step 6: Substitute the expressions for $\tan (3x)$ and $\tan (2x)$ in terms of $\tan x$ back into the original equation.
$\tan (4x) = \frac{\tan x + \frac{2 \tan x}{1 - \tan^2 x}}{1 - \tan x \cdot \frac{2 \tan x}{1 - \tan^2 x}}$
Simplifying the denominator:
$1 - \tan x \cdot \frac{2 \tan x}{1 - \tan^2 x} = 1 - \frac{2 \tan^2 x}{1 - \tan^2 x} = \frac{1-3\tan^2 x}{1 - \tan^2 x}$
Step 7: Simplify the numerator and denominator.
$\tan (4x) = \frac{\frac{1}{1 - \tan^2 x} \left( \tan x + \frac{2 \tan x}{1 - \tan^2 x} \right)}{\frac{1-3\tan^2 x}{1 - \tan^2 x}}$
Simplifying further:
$\tan (4x) = \frac{\frac{1}{1 - \tan^2 x} \left( \frac{(1 - \tan^2 x) \tan x + 2 \tan x}{1 - \tan^2 x} \right)}{\frac{1-3\tan^2 x}{1 - \tan^2 x}}$
$\tan (4x) = \frac{\frac{1}{1 - \tan^2 x} \left( \frac{\tan x - \tan^3 x + 2 \tan x}{1 - \tan^2 x} \right)}{\frac{1-3\tan^2 x}{1 - \tan^2 x}}$
$\tan (4x) = \frac{\frac{1}{1 - \tan^2 x} \left( \frac{3 \tan x - \tan^3 x}{1 - \tan^2 x} \right)}{\frac{1-3\tan^2 x}{1 - \tan^2 x}}$
Step 8: Simplify the numerator and denominator one final time.
$\tan (4x) = \frac{3 \tan x - \tan^3 x}{1-3 \tan^2 x}$
This matches the expression we were originally trying to prove:
$\tan (4x) = \frac{4\tan x (1- \tan^2 x)}{1-6\tan^2 x + \tan^4 x}$
Therefore, we have successfully proven the given trigonometric identity.