Question

Let $0<\theta <\pi /2$ and
$\Delta (x,\theta )=\left|\begin{array}{ccc}x& \tan \theta & \cot \theta \\ -\tan \theta & -x& 1\\ \cot \theta & 1& x\end{array}\right|$
then

• A. $\Delta (0,\theta )=0$
• B. $\Delta (x,\frac{\pi }{4})=x-x^3$
• C. $Mi{n}_{0<\theta <\pi /2}\Delta (1,\theta )=0$
• D. $\Delta (x,\theta )$ is independent of x

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\Delta (x,\frac{\pi }{4})=x-x^3$
B $\Delta (0,\theta )=0$
C $Mi{n}_{0<\theta <\pi /2}\Delta (1,\theta )=0$

$\Delta (0,\theta )=\left|\begin{array}{ccc}0& \tan \theta & \cot \theta \\ -\tan \theta & 0& 1\\ \cot \theta & 1& 0\end{array}\right|=\tan \theta \cot \theta -\cot \theta \tan \theta =0$
$\Delta (x,\pi /4)=\left|\begin{array}{ccc}x& 1& 1\\ -1& -x& 1\\ 1& 1& x\end{array}\right|=x(-x^2-1)+1(1+x)+1(x-1)$

$=-x^3-x+1+x+x-1=x-x^3$
$\Delta (1,\theta )=\left|\begin{array}{ccc}1& \tan \theta & \cot \theta \\ -\tan \theta & -1& 1\\ \cot \theta & 1& 1\end{array}\right|=-2+\tan \theta (\cot \theta +\tan \theta )+\cot \theta (\cot \theta -\tan \theta )=-2+{\tan }^2\theta +{\cot }^2\theta$
${\tan }^2\theta +{\cot }^2\theta \ge 2$

Hence,

$\Delta {(1,\theta )}_{min.}=0$