Question

If $\left|\begin{array}{ccc}1& \sin x& {\sin }^{2}x\\ 1& \cos x& {\cos }^{2}x\\ 1& \tan x& {\tan }^{2}x\end{array}\right|=0,x\in [0,2\pi ],$ then number of possible values of $x$ is.

• A. $4$
• B. $5$
• C. $6$
• D. None of these

Answer 1

The correct option is D None of these

$\left|\stackrel{1}{\underset{1}{1}}\stackrel{sinx}{\underset{tanx}{cosx}}\stackrel{{\sin }^{2}x}{\underset{{\tan }^{2}x}{{\cos }^{2}x}}\right|=0$

$\left(\underset{R_2\to R_2-R_3}{R_1\to R_1-R_2}\right)$

$\Rightarrow \left|\stackrel{0}{\underset{1}{0}}\stackrel{sinx-cosx}{\underset{tanx}{cosx-tanx}}\stackrel{{\sin }^{2}x-{\cos }^{2}x}{\underset{{\tan }^{2}x}{{\cos }^{2}x-{\tan }^{2}x}}\right|=0$

$\Rightarrow \underset{(cosx-tanx)}{(sinx-cosx)}\left|\stackrel{0}{\underset{1}{0}}\stackrel{1}{\underset{tanx}{1}}\stackrel{sinx+cosx}{\underset{{\tan }^{2}x}{cosx+tanx}}\right|=0$

Expanding along colomn I we get

$\Rightarrow (cosx-tanx)(sinx-cosx)(cosx+tanx-sinx-cosx)=0$

$\Rightarrow (cosx-tanx)(sinx-cosx)(tanx-sinx)=0$

$\therefore$ $cosx-tanx=0$ or $sinx=cosx$ or $tanx=sinx$

or $cosx=tanx$ ; $sinx=cosx$ ; $tanx=sinx$

for $xϵ[0,2\pi ]$

$csox=tanx$

from graph we see there are $2$ solutions.

for $xϵ[0,2\pi ]$

$sinx=cosx$

from graph we have two solutions

for $xϵ[0,2\pi ]$

from graph we have solutions

So a total of $2+2+3=7$ solutions

i.e Answer : D.