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Question: If A, B, C be the angles of a triangle, then \(\left| \begin{matrix} - 1 & \cos C & \cos B \\ \cos ...

If A, B, C be the angles of a triangle, then $\left| \begin{matrix}

  • 1 & \cos C & \cos B \ \cos C & - 1 & \cos A \ \cos B & \cos A & - 1 \end{matrix} \right| =$
A

1

B

0

C

cos⁡Acos⁡Bcos⁡C\cos A\cos B\cos C

D

cos⁡A+cos⁡Bcos⁡C\cos A + \cos B\cos C

Answer

0

Explanation

Solution

Given, Angles of a triangle = A, B and C. We know that as

A + B + C = π, therefore A+B=π−CA + B = \pi - C

or cos⁡(A+B)=cos⁡(π−C)=−cos⁡C\cos(A + B) = \cos(\pi - C) = - \cos C

or cos⁡Acos⁡B−sin⁡Asin⁡B=−cos⁡C\cos A\cos B - \sin A\sin B = - \cos C

cos⁡Acos⁡B+cos⁡C=sin⁡Asin⁡B\cos A\cos B + \cos C = \sin A\sin B

and sin⁡(A+B)=sin⁡(π−C)=sin⁡C.\sin(A + B) = \sin(\pi - C) = \sin C.

Expanding the given determinant, we get

Δ=−(1−cos⁡2A)+cos⁡C(cos⁡C+cos⁡Acos⁡B)+cos⁡B(cos⁡B+cos⁡Acos⁡C)\Delta = - (1 - \cos^{2}A) + \cos C(\cos C + \cos A\cos B) + \cos B(\cos B + \cos A\cos C)

=−sin⁡2A+cos⁡C(sin⁡Asin⁡B)+cos⁡B(sin⁡Asin⁡C)= - \sin^{2}A + \cos C(\sin A\sin B) + \cos B(\sin A\sin C)

=−sin⁡2A+sin⁡A(sin⁡Bcos⁡C+cos⁡Bsin⁡C)= - \sin^{2}A + \sin A(\sin B\cos C + \cos B\sin C)

=−sin⁡2A+sin⁡Asin⁡(B+C)=−sin⁡2A+sin⁡2A=0.= - \sin^{2}A + \sin A\sin(B + C) = - \sin^{2}A + \sin^{2}A = 0.