If A- B- C be the angles of a triangle- then prove that - -1 cosC cosB- cosC -1 cosA- cosB cosA -1- - - - 0

In a triangle #160;cos(A+B) = cos (π C) = cosC #160; #160; ....(1) ∴ cosAcosB+cosC= sinAsinB etc.sin(A+B) = sinC etc.Expanding the given determina ...

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Standard XII

Mathematics

Properties of Determinants

If A, B, C be...

Question

If A, B, C be the angles of a triangle, then prove that
$∣ ∣ ∣ ∣ \begin{matrix} - 1 & c o s C & c o s B c o s C & - 1 & c o s A c o s B & c o s A & - 1 \end{matrix} ∣ ∣ ∣ ∣$ = 0

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Solution

In a triangle
cos(A+B) = cos (\pi-C) = - cosC ....(1)
$∴$ cosAcosB+cosC= sinAsinB etc.
sin(A+B) = sinC etc.
Expanding the given determinant, we get
$△ = - (1 - c o s^{2} A) + c o s C [c o s C + c o s A c o s B] + c o s B [c o s B + c o s A c o s C]$
$= - s i n^{2} A + c o s C (s i n A s i n B) + c o s B (s i n A s i n C)$
= $- s i n^{2} A + s i n A s i n (B + C)$
= $- s i n^{2} A + s i n^{2} A = 0.$

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If A, B, C be the angles of a triangle, then prove that ∣∣ ∣∣−1cosCcosBcosC−1cosAcosBcosA−1∣∣ ∣∣ = 0

In a triangle cos(A+B) = cos (\pi-C) = - cosC ....(1)∴ cosAcosB+cosC= sinAsinB etc.sin(A+B) = sinC etc.Expanding the given determinant, we get△=−(1−cos2A)+cosC[cosC+cosAcosB]+cosB[cosB+cosAcosC]=−sin2A+cosC(sinAsinB)+cosB(sinAsinC)=−sin2A+sinAsin(B+C)=−sin2A+sin2A=0.

If A, B, C be the angles of a triangle, then prove that
$∣ ∣ ∣ ∣ \begin{matrix} - 1 & c o s C & c o s B c o s C & - 1 & c o s A c o s B & c o s A & - 1 \end{matrix} ∣ ∣ ∣ ∣$ = 0