Question

The number of integers in the range of $\frac{13-{\tan }^{2}x}{1+{\tan }^{2}x}$ is

Answer 1

Given: $\frac{13-{\tan }^{2}x}{1+{\tan }^{2}x}$
We know that,
$x\ne (2n+1)\frac{\pi }{2}$
Now,
$\frac{13-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1-{\tan }^{2}x+12}{1+{\tan }^{2}x}=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}+\frac{12}{1+{\tan }^{2}x}=\cos 2x+12{\cos }^{2}x=\cos 2x+6(1+\cos 2x)=7\cos 2x+6$

As $x\ne (2n+1)\frac{\pi }{2}$, so
$\cos 2x\in (-1,1]$
Therefore,
$\frac{13-{\tan }^{2}x}{1+{\tan }^{2}x}\in (-1,13]$

Hence, the number of integral values in the range is $14$.