Question

$A+B = C\\Prove that\\cos^{2}A+cos^{2}B+cos^{2}C = 1+2cosAcosBcosC$

Answer 1

Correct question:-

If A+B+C=π then,Prove that: cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1

Solution:-

Here, A + B + C = π, then

A + B = π - C

B + C = π - A

A + C = π - C

$\bigstar$ Now, Let’s consider:-

= 2 cosA . cosB . cosC

= cosA [2 cosB.cosC]

= cosA[cos(B+C)+ cos(B- C)]

= cosA [cos(180-A) + cos(B-C)]

= cosA [-cosA + cos(B-C)]

= -cos²A + cosA cos(B-C)

= -cos²A + {2cosA cos(B-C)}/2

= -cos²A + [cos(A - B + C) + cos(A + B - C)]/ 2

= -cos²A + [cos(A + C - B) + cos(A + B - C)]/ 2

= -cos²A + [cos(180 - B - B) + cos(180 - C - C)]/2

= -cos²A + [cos(180 - 2B) + cos(180 - 2C)]/2

= -cos²A – [cos2B + cos2C]/2

= ( -2cos²A – cos2B – cos2C ) / 2

= 1/2 [-2cos²A – cos2B – cos2C]

= 1/2 [ -2cos²A – 2cos²B + 1 – 2cos²C + 1 ]

=1/2 [ -2cos²A – 2cos²B – 2cos²C + 2 ]

= -(cos²A + cos²B + cos²C) + 1

or 2 cosA . cosB . cosC= – (cos²A + cos²B + cos²C) + 1

cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1

Hence Proved!

Answer 2

Answer:

Solution,

cos(A+B) = cos C

L.HS = $cos^2A +cos^2B + cos^2C$

       =$(1/2)(2cos^2A +2cos^2B+2cos^2C)$

       =$\frac{1}{2}*(cos2A+1+cos2B+1+2cos^2C)$

       =$\frac{1}{2}(2+cos2A+cos2B+2cos^2C)$

       =$\frac{1}{2}[2+(2cos(A+B).cos(A-B))+2cos^2C]$

       =$\frac{1}{2}[2+2.cosC.cos(A-B)+2cos^2C]$

       =$\frac{1}{2}[2+2.cosC.{cos(A-B)+cos(A+B)}]$

       =$1+2cosAcosBcosC$

       =RHS

Hope it will help you a lot