Question

If $\tan \left[\frac{\pi }{4}+\theta \right]+\tan \left[\frac{\pi }{4}-\theta \right]=\alpha$ then ${\tan }^3\left[\frac{\pi }{4}+\theta \right]+{\tan }^3\left[\frac{\pi }{4}-\theta \right]=$

Answer 1

We have,

$\tan \left[\frac{\pi }{4}+\theta \right]+\tan \left[\frac{\pi }{4}-\theta \right]=\alpha$ $............\left(1\right)$

On squaring both sides, we get

${[\tan \left[\frac{\pi }{4}+\theta \right]+\tan \left[\frac{\pi }{4}-\theta \right]]}^2={\alpha }^2$

${\tan }^2\left[\frac{\pi }{4}+\theta \right]+{\tan }^2\left[\frac{\pi }{4}-\theta \right]+2\tan [\frac{\pi }{4}+\theta ]\tan [\frac{\pi }{4}-\theta ]={\alpha }^2$

Since,

$\tan [\frac{\pi }{4}+\theta ]\tan [\frac{\pi }{4}-\theta ]=1$

Therefore,

${\tan }^2\left[\frac{\pi }{4}+\theta \right]+{\tan }^2\left[\frac{\pi }{4}-\theta \right]+2={\alpha }^2$

${\tan }^2\left[\frac{\pi }{4}+\theta \right]+{\tan }^2\left[\frac{\pi }{4}-\theta \right]={\alpha }^2-2$ $...........\left(2\right)$

Since,

${\tan }^3\left[\frac{\pi }{4}+\theta \right]+{\tan }^3\left[\frac{\pi }{4}-\theta \right]$

$\Rightarrow (\tan \left[\frac{\pi }{4}+\theta \right]+\tan \left[\frac{\pi }{4}-\theta \right])({\tan }^2[\frac{\pi }{4}+\theta ]+{\tan }^2[\frac{\pi }{4}-\theta ]-\tan [\frac{\pi }{4}+\theta ]\tan [\frac{\pi }{4}-\theta ])$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow \left(\alpha \right)({\alpha }^2-2-1)$

$\Rightarrow \alpha ({\alpha }^2-3)$

Hence, this is the answer.