Question

The maximum value of $f\left(x\right)=\frac{{\tan }^{2}x-{\cot }^{2}x+1}{{\tan }^{2}x+{\cot }^{2}x-1}$ is

• A. $\frac{3}{2}$
• B. $3$
• C. $\frac{5}{2}$
• D. None of these

Answer 1

Answer: (D)

None of these

$f\left(x\right)=\frac{{\tan }^{2}x-{\cot }^{2}x+1}{{\tan }^{2}x+{\cot }^{2}x-1}$
Let $t={\tan }^{2}x$ and $f\left(x\right)=y$
Thus, $y=\frac{t^2+t-1}{t^2-t+1}$
$\Rightarrow y{t}^2-ty+y-t^2-t+1=0$
$\Rightarrow t^2(y-1)+t(-y-1)+y+1=0$
$\Rightarrow (y+1{)}^2-4(y-1\left)\right(y+1)>0$ since t is real.
$\Rightarrow (y+1)(y+1-4y+4)>0$ or $(y+1)(3y-5)<0$
$\Rightarrow y\in (-1,\frac{5}{3})$