Prove the following:

In ∆ABC, ∠C = `(2pi)/3`, then prove that cos^{2}A + cos^{2}B − cos A cos B = `3/4`

#### Solution

In ∆ABC, A + B + C = π, where ∠C = `(2pi)/3`

∴ A + B = π – C = `pi/3` ...(1)

L.H.S. = cos^{2}A + cos^{2}B – cos A cos B

= cos^{2}A + 1 – sin^{2}B – cos A cos B

= 1 + (cos^{2}A – sin^{2}B) – cos A cos B

= 1 + cos (A + B) · cos (A – B) – cos A cos B ...[∵ cos (A + B) · cos (A – B) = cos^{2}A – sin^{2}B]

= `1 + cos pi/3cos("A" - "B") - cos"A" cos"B"` ...[By (1)]

= `1 + 1/2cos("A" - "B") - cos"A" cos"B"`

= `1 + 1/2(cos"A" cos"B" + sin"A" sin"B") - cos"A" cos"B"`

= `1 + 1/2cos"A"cos"B" + 1/2sin "A" sin"B" - cos"A"cos"B"`

= `1 + 1/2sin"A"sin"B" - 1/2cos"A"cos"B"`

= `1 - 1/2(cos"A" cos"B" - sin"A"sin"B")`

= `1 - 1/2cos("A" + "B")`

= `1 - 1/2cos pi/3` ...[By (1)]

= `1 - 1/2 xx 1/2`

= `1 - 1/4`

= `3/4`

= R.H.S.