अंतराल -frac{pi}{4} leq x leq frac{pi}{4} में | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\left|\begin{array}{lll} \sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x \end{array}ight|=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$\left|\begin{array}{lll} \sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x \end{array}ight|=0$, $\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$

Apply: $R_{1} ightarrow R_{1}-R_{2}$ and $R_{2} ightarrow R_{2}-R_{3}$

$\left|\begin{array}{ccc} \sin x-\cos x & \cos x-\sin x & 0 \ 0 & \sin x-\cos x & \cos x-\sin x \ \cos x & \cos x & \sin x \end{array}ight|=0$

$(\sin x-\cos x)^{2}\left|\begin{array}{ccc}1 & -1 & 0 \ 0 & 1 & -1 \ \cos x & \cos x & \sin x\end{array}ight|=0$

$(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$

$\sin x=\cos x \quad$ or $\quad \sin x=-2 \cos x$

$	an x=1$ or $\quad 	an x=-2$ (Not valid)

$	herefore x=\frac{\pi}{4}$

अंतराल $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ में, $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ के भिन्न वास्तविक मूलों की संख्या है

Similar Questions

सारणिकों का मान ज्ञात कीजिए:

$\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

यदि समीकरणों के निकाय $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ का कोई हल नहीं है, तब $\alpha $ का मान है

यदि $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, तो $k =$

$\left| {\,\begin{array}{*{20}{c}}{{{\sin }^2}x}&{{{\cos }^2}x}&1\\{{{\cos }^2}x}&{{{\sin }^2}x}&1\\{ - 10}&{12}&2\end{array}\,} \right| = $