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If $1+\sin x+\cos x\neq 0$ the value of $x$ for which $\begin{vmatrix}1&\sin x&\cos x\\\sin x&1&\cos x\\\cos x&\sin x&1\end{vmatrix}=0$ is

$\begin{array}{1 1}(A)\;n\pi+(-1)^n\large\frac{\pi}{2},\normalsize 2n\pi&(B)\;n\pi,2n\pi\pm \large\frac{\pi}{2}&\\(C)\;n\pi+(-1)^n\large\frac{\pi}{2},\normalsize 2n\pi\pm \large\frac{\pi}{2}&(D)\;\text{None of these}\end{array}$

1 Answer

$\begin{vmatrix}1&\sin x&\cos x\\\sin x&1&\cos x\\\cos x&\sin x&1\end{vmatrix}=0$
Applying $C_1+C_2+C_3$
$\Delta=\begin{vmatrix}1+\sin x+\cos x&\sin x&\cos x\\1+\sin x+\cos x&1&\cos x\\1+\sin x+\cos x&\sin x&1\end{vmatrix}$
Taking out $1+\sin x+\cos x$ we get
$\Delta=1+\sin x+\cos x\begin{vmatrix}1&\sin x&\cos x\\1&1&\cos x\\1&\sin x&1\end{vmatrix}$
Applying $R_1-R_2=R_1$
$(1+\sin x+\cos x)\begin{vmatrix}0&\sin x-1&0\\1&1&\cos x\\1&\sin x&1\end{vmatrix}$
$\Rightarrow (1+\sin x+\cos x)(\sin x-1)(1-\cos x)$
$\Delta=0\Rightarrow \sin x=1$
$\Rightarrow \cos x=1$
$\Rightarrow x=n\pi+(-1)^n\large\frac{\pi}{2}$$,2n\pi$
Hence (A) is the correct answer.
answered Apr 25, 2014 by sreemathi.v

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