Question

Find the value of $\theta$ laying between 0 and $\frac{\pi }{2}$ and satisfying the equation

$\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\sin 4\theta \end{array}\right|=0$.

If the value of $\theta =\frac{a\pi }{b}$, then find the value of $\frac{(b-3)}{a}$

Answer 1

$\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\sin 4\theta \end{array}\right|=0$
$R_1\to R_1-R_2,R_3\to R_3-R_2$
$\left|\begin{array}{ccc}1& -1& 0\\ co{s}^2\theta & 1+si{n}^2\theta & 4sin4\theta \\ 0& -1& 1\end{array}\right|=0$
$C_2\to C_2+C_1$
$\left|\begin{array}{ccc}1& 0& 0\\ co{s}^2\theta & 2& 4sin4\theta \\ 0& -1& 1\end{array}\right|=0$
$\Rightarrow 2+4\sin \theta =0$
$\Rightarrow \sin 4\theta =-\frac{1}{2}$
$\Rightarrow \sin 4\theta =-\sin \left(\frac{\pi }{6}\right)$
$\Rightarrow \sin 4\theta =\sin (\pi +\frac{\pi }{6})$
$\Rightarrow 4\theta =\frac{7\pi }{6}$
$\Rightarrow \theta =\frac{7\pi }{24}$

Comparing with the given value of $\theta$, we get

$a=7,b=24$

Value of $\frac{b-3}{a}=3$