જો $ \text{A, B, C}$ એ ત્રિકોણના ખૂણા હોય , તો $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}\,} \right| = $
A$1$
B$0$
C$\cos A\cos B\cos C$
D$\cos A + \cos B\cos C$
Difficult
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B$0$
Given, Angles of a triangle $= A, B$ and $C$.
We know that as $A + B + C = \pi ,$
therefore $A + B = \pi - C$
or $\cos (A + B) = \cos (\pi - C) = - \cos C$
or $\cos A\cos B - \sin A\sin B = - \cos C$
$\cos A\cos B + \cos C = \sin A\sin B$
and $\sin (A + B) = \sin (\pi - C) = \sin C.$
Expanding the given determinant, we get
$\Delta = - (1 - {\cos ^2}A) + \cos C(\cos C + \cos A\cos B) + \cos B(\cos B + \cos A\cos C)$
$ = - {\sin ^2}A + \cos C(\sin A\sin B) + \cos B(\sin A\sin C)$
$ = - {\sin ^2}A + \sin A(\sin B\cos C + \cos B\sin C)$
$ = - {\sin ^2}A + \sin A\sin (B + C)$
$ = - {\sin ^2}A + {\sin ^2}A = 0.$
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જો $f(x)=\left|\begin{array}{ccc}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x\end{array}\right|$ હોય, તો $\frac{1}{5} f^{\prime}(0)$ ____________
જો $'a'$ એ અવાસ્તવિક સંકર સંખ્યા છે કે જેથી સમીકરણો $ax -a^2y + a^3z= 0$ , $-a^2x + a^3y + az = 0$ અને $a^3x + ay -a^2z = 0$ ને શૂન્યતર ઉકેલ હોય તો $|a|$ મેળવો.
ધારો કે $x , y , z > 1$ અને $A=\left[\begin{array}{lll}1 & \log _x y & \log _x z \\ \log _y x & 2 & \log _y z \\ \log _z x & \log _z y & 3\end{array}\right]$ તો $\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right| =.........$
જો $a,b,c$ અને $d$ એ સંકર સંખ્યા હોય , તો નિશ્રાયક $\Delta = \left| {\,\begin{array}{*{20}{c}}2&{a + b + c + d}&{ab + cd}\\{a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)}\\{ab + cd}&{ab(c + d) + cd(a + d)}&{2abcd}\end{array}} \right|$ એ $. . ..$ પર આધારિત છે.