Question

If $x,yϵR$, then the determinant $\Delta =\left|\begin{array}{ccc}cosx& -sinx& 1\\ sinx& cosx& 1\\ cos(x+y)& -sin(x+y)& 0\end{array}\right|$ lies in the interval

• A. $[-\sqrt{2},\sqrt{2}]$
  
• B. [-1, 1]
• C. $[-\sqrt{2},1]$
• D. $[-1,-\sqrt{2}]$

Answer 1

The correct option is B [-1, 1]

$\Delta =\left|\begin{array}{ccc}cosx& -sinx& 1\\ sinx& cosx& 1\\ cos(x+y)& -sin(x+y)& 0\end{array}\right|R_2=R_1-R_2\left|\begin{array}{ccc}cosx& -sinx& 1\\ cosx-sinx& -(sinx+cosx)& 0\\ cos(x+y)& -sin(x+y)& 0\end{array}\right|$
$=-1\left|\begin{array}{ccc}cosx& -sinx& 1\\ cosx-sinx& sinx+cosx& 0\\ cos(x+y)& sin(x+y)& 0\end{array}\right|R_2=R_1-R_2\left|\begin{array}{ccc}cosx& -sinx& 1\\ sinx& -cosx& 0\\ cos(x+y)& -sin(x+y)& 0\end{array}\right|$
$=-1[sinx.sin(x+y)+cosx(cos(x+y)\left)\right]$
$=-1[si{n}^{2}xcosy+\text{/}sinx.cosxsiny+co{s}^{2}x.cosy-\text{/}cosx.sinx.siny]$
$=-cosy(si{n}^{2}x+co{s}^{2}x)=-cosy$
$\Rightarrow$ Range of $\Delta =[-1,1]$