Question

If $\Delta (x)\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|,$ then $\underset{0}{\overset{\pi /4}{\int }}\Delta \left(x\right)dx$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 0
• D. $-\frac{1}{4}$

Answer 1

Answer: (D)

$\Delta (x)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|,$
Applying $C_3\to C_3+C_2-C_1$
$\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 0\\ 1+\sin x& \cos x& 0\\ \sin x& \sin x& 1\end{array}\right|$
$=\cos x-\cos x\left(1+\sin x\right)$
$[\because$ expanding along $C_3$]
$=-\cos x.\sin x=-\frac{1}{2}\sin 2x$
$\therefore \underset{0}{\overset{\pi /4}{\int }}\Delta \left(x\right)dx$
$=-\frac{1}{2}\underset{0}{\overset{\pi /4}{\int }}\sin 2xdx$
$=-\frac{1}{2}{\left[-\frac{\cos 2x}{2}\right]}_0^{\pi /4}$
$=+\frac{1}{2\times 2}\left[\cos \frac{\pi }{2}-\cos 0^{\circ }\right]$
$=\frac{1}{4}\left(0-1\right)=-\frac{1}{4}$