If P, Q and R are angles of Delta PQR, then t | vedclass

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(B) Let $\Delta = \left|\begin{array}{ccc}-1 & \cos R & \cos Q \ \cos R & -1 & \cos P \ \cos Q & \cos P & -1\end{array}ight|$.Expanding the determ

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(B) Let $\Delta = \left|\begin{array}{ccc}-1 & \cos R & \cos Q \ \cos R & -1 & \cos P \ \cos Q & \cos P & -1\end{array}ight|$.Expanding the determinant along the first row:$\Delta = -1(1 - \cos^2 P) - \cos R(-\cos R - \cos Q \cos P) + \cos Q(\cos R \cos P + \cos Q)$$\Delta = -(1 - \cos^2 P) + \cos^2 R + \cos R \cos Q \cos P + \cos Q \cos R \cos P + \cos^2 Q$$\Delta = -1 + \cos^2 P + \cos^2 R + \cos^2 Q + 2 \cos P \cos Q \cos R$.Since $P, Q, R$ are angles of a triangle, $P+Q+R = \pi$, so $R = \pi - (P+Q)$.Using the identity for the sum of squares of cosines in a triangle: $\cos^2 P + \cos^2 Q + \cos^2 R = 1 - 2 \cos P \cos Q \cos R$.Substituting this into the expression for $\Delta$:$\Delta = -1 + (1 - 2 \cos P \cos Q \cos R) + 2 \cos P \cos Q \cos R = 0$.

If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

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