Question

Which of the following options for the equation ${\tan }^{2}x-\sqrt{5}\tan x+1=0$ in $0<x<\pi /2$ is correct?

• A. no solution exists
• B. two solutions $x_1$ and $x_2$ exists with $x_1+x_2=\pi /4$
• C. two solutions $x_1$ and $x_2$ exists with $x_1+x_2=\pi /2$
• D. two solutions $x_1$ and $x_2$ exists with $x_1+x_2=\pi /6$

Answer 1

The correct option is C two solutions $x_1$ and $x_2$ exists with $x_1+x_2=\pi /2$

${\tan }^{2}x-\sqrt{5}\tan x+1=0$
Let $\tan x_1,\tan x_2$ be the roots of the equation, then
$\tan x_1+\tan x_2=\sqrt{5}\tan x_1\tan x_2=1$

Now,
$\tan (x_1+x_2)=\frac{\tan x_1+\tan x_2}{1-\tan x_1\tan x_2}\Rightarrow \tan (x_1+x_2)=\frac{\sqrt{5}}{1-1}\therefore x_1+x_2=\frac{\pi }{2}$