Question

If $0^0\le x\le 90$ and $1+{\tan }^{2}x-2\tan x=0$, then the value of $x$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

Answer: (B)

${45}^0$

We have $0^0\le x\le 90$

Given, $1+{\tan }^{2}x-2\tan x=0$
It can be written as ${(1-\tan x)}^2=0$
$\Rightarrow \tan x=1$

$\Rightarrow x={45}^0$