left|begin{array}{ccc}cos alpha cos beta & co | vedclass

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(C) ધારો કે $\Delta = \left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \

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(C) ધારો કે $\Delta = \left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}ight|$.ત્રીજા સ્તંભ $(C_3)$ ની સાપેક્ષ વિસ્તરણ કરતા:$\Delta = -\sin \alpha \left( (- \sin \beta)(\sin \alpha \sin \beta) - (\cos \beta)(\sin \alpha \cos \beta) ight) + \cos \alpha \left( (\cos \alpha \cos \beta)(\cos \beta) - (\cos \alpha \sin \beta)(-\sin \beta) ight)$$\Delta = -\sin \alpha \left( -\sin \alpha \sin^2 \beta - \sin \alpha \cos^2 \beta ight) + \cos \alpha \left( \cos \alpha \cos^2 \beta + \cos \alpha \sin^2 \beta ight)$$\Delta = -\sin \alpha [-\sin \alpha (\sin^2 \beta + \cos^2 \beta)] + \cos \alpha [\cos \alpha (\cos^2 \beta + \sin^2 \beta)]$કારણ કે $\sin^2 \beta + \cos^2 \beta = 1$,તેથી:$\Delta = -\sin \alpha [-\sin \alpha (1)] + \cos \alpha [\cos \alpha (1)]$$\Delta = \sin^2 \alpha + \cos^2 \alpha$$\Delta = 1$.

$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$ ની કિંમત શોધો.

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$\left| {\begin{array}{ccc} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{array}} \right| = $

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