Question

If $△=\left|\begin{array}{ccc}\sin \pi & \cos (x+\frac{\pi }{4})& \tan (x-\frac{\pi }{4})\\ \sin (x-\frac{\pi }{4})& 0& \ln \left(\frac{x}{y}\right)\\ \cot (x+\frac{\pi }{4})& \ln \left(\frac{y}{x}\right)& 0\end{array}\right|,$ for $x\in (0,\pi )-\{\frac{\pi }{4},\frac{3\pi }{4}\},y>0.$ Then the value of $(△+9)=$

Answer 1

Using : $\cos (x+\frac{\pi }{4})=\sin [\frac{\pi }{2}-(x+\frac{\pi }{4})]=\sin (\frac{\pi }{4}-x)=-\sin (x-\frac{\pi }{4})$
and $\cot (x+\frac{\pi }{4})=\tan [\frac{\pi }{2}-(\frac{\pi }{4}+x)]=\tan (\frac{\pi }{4}-x)=-\tan (x-\frac{\pi }{4})$
and $\ln \left(\frac{x}{y}\right)=-\ln \left(\frac{y}{x}\right)$
So, determinant becomes :
$△=\left|\begin{array}{ccc}0& -\sin (x-\frac{\pi }{4})& \tan (x-\frac{\pi }{4})\\ \sin (x-\frac{\pi }{4})& 0& -\ln \left(\frac{y}{x}\right)\\ -\tan (x-\frac{\pi }{4})& \ln \left(\frac{y}{x}\right)& 0\end{array}\right|$
Which is determinant of odd order skew-symmetric matrix.
$\Rightarrow △=0$