Question

The number of solution of $|\tan x|=tanx+\frac{1}{\cos x}$ in $[0,2\pi ]$ is

• A. $4$
• B. $1$
• C. $2$
• D. $6$

Answer 1

The correct option is B $1$

Let split domain in two parts $[0,\frac{\pi }{2}]\cup [\pi ,\frac{3\pi }{2}]$$and[\frac{\pi }{2},\pi ]\cup [\frac{3\pi }{2},2\pi ]$
Case - 1
$xϵ[0,\frac{\pi }{2}]\cup [\pi ,\frac{3\pi }{2}],|tanx|=tanx=tanx+\frac{1}{cosx}$
$\frac{1}{cosx}=0$
$\Rightarrow$ no solution.
Case - 2
$xϵ[\frac{\pi }{2},\pi ]\cup [\frac{3\pi }{2},2\pi ]$

$|tanx|=-tanx=tanx+\frac{1}{cosx}$
$0=\frac{2sinx+1}{cosx}$
$\Rightarrow sinx=-\frac{1}{2}$

only one solution i.e., $x={330}^o$