Question

Number of solutions of $1+{\tan }^{2}x-\sec x\tan x<0$ is

• A. 0
• B. 2
• C. 3
• D. infinite

Answer 1

The correct option is D infinite

We know that $1+ta{n}^{2}x=se{c}^{2}x$
So, given inequation is $se{c}^{2}x-secxtanx<0$
$=>secx(secx-tanx)<0$
$=>\frac{1}{cosx}(\frac{1}{cosx}-\frac{sinx}{cosx})<0$
$=>\frac{1-sinx}{co{s}^{2}x}<0$
$=>\frac{1-sinx}{1-si{n}^{2}x}<0$
$=>\frac{1-sinx}{(1-sinx)(1+sinx)}<0$
$=>\frac{1}{1+sinx}<0$
$=>1+sinx>0$
$=>sinx>-1$
There are many values of $x$ which satisfy the inequation.