Question

If 0 < $\alpha <\frac{\pi }{4(n-1)}$ where n > 1 , then prove that $\tan n$ $\alpha >n\tan \alpha$

Answer 1

Since $n>1,p\left(2\right)$ holds as
$\tan$ 2$\alpha$ > $2\tan$ $\alpha$
when 0 < $\alpha$ $<\frac{\pi }{4}$$(n=2)$
$\therefore \frac{2\tan \alpha }{1-{\tan }^2\alpha }$ 2 tan $\alpha$ as tan $\alpha$ is +ive and < 1
Now assume P (m) i.e., $\tan$$\alpha$ > $m\tan$ $\alpha$
Now $\tan (m+1)$$\alpha$ = $\frac{\tan m\alpha +\tan \alpha }{1-\tan m\alpha \tan \alpha }$ ...(1)
Where $0<\alpha <\frac{\pi }{4(m+1-1)}$
or $0<\alpha <\frac{\pi }{4m}or0<m\alpha <\frac{\pi }{4}$
$\therefore \tan m\alpha <1$ and is $+ive$ . Also $\tan$$\alpha$ is < 1 and +ive .
Hence from (1)
$\tan (m+1)$ $\alpha$ > $m\tan$ $\alpha$ $+\tan$ $\alpha$ $=(m+1)\tan$ $\alpha$
$\therefore$ $P(m+1)$ also holds good .
Hence universally true.