Question

prove that ,a^2(cos^2B-cos^2C)+b^2(cos^2C-cos^2A)+c^2(cos^2A-cos^2B)=0​

Answer 1

Step-by-step explanation:

To prove--->

a²(cos²B - Cos²C)+b²(Cos²C- Cos²A)+c²(Cos²A-Cos²B) = 0

Proof---> We have some formula

(1) Sine Rule

a / SinA = b / SinB = C / SinC = k

Now a / SinA = k

=> a = k SinA

Similarly,

b = k SinB , c = k SinC

(2) Sin²θ + Cos²θ = 1

Cos²θ = 1 - Sin²θ

Now returning to original problem . first we solve the first term of LHS

First term = a² ( Cos²B - Cos²C )

= ( kSinA )² { 1 - Sin²B - ( 1 - Sin² C) }

= k² Sin²A ( 1 - Sin²B - 1 + Sin²C )

= k² Sin²A ( Sin²C - Sin²B )

= k² ( Sin²A Sin²C - Sin²B Sin²A )

= k²Sin²A Sin²C - k²Sin²B Sin²A

Similarly,

Second term = k² Sin²B Sin²A - k² Sin²C Sin²B

Third term = k²Sin²C Sin²B - k² Sin² A Sin²C )

LHS = first term + Second term + third term

= K²Sin²ASin²C - k² Sin²BSin²A+ k²Sin²BSin²A-k² Sin²CSin²B + k²Sin²CSin²B - k²Sin²ASin²C

Clearly all terms cancel out each other

= 0 = RHS