For any Δ ABC show that c(a cos B – b cos A) = a2 – b2

1.	For any Δ ABC show that c(a cos B – b cos A) = a2 – b2
Answer» Let us consider the LHS c(a cos B – b cos A) Now, as LHS contain ca cos B and cb cos A which can be obtained from cosine formulae. Then, from cosine formula we have Cos A = (b² + c² – a²)/2bc bc cos A = (b² + c² – a²)/2 … (i) Cos B = (a² + c² – b²)/2ac ac cos B = (a² + c² – b²)/2 … (ii) Let us subtract equation (ii) from (i) we get, ac cos B – bc cos A = (a² + c² – b²)/2 – (b² + c² – a²)/2 = a² – b² ∴ c(a cos B – b cos A) = a² – b² Thus proved.

Answer»

Let us consider the LHS

c(a cos B – b cos A)

Now, as LHS contain ca cos B and cb cos A which can be obtained from cosine formulae.

Then, from cosine formula we have

Cos A = (b² + c² – a²)/2bc

bc cos A = (b² + c² – a²)/2 … (i)

Cos B = (a² + c² – b²)/2ac

ac cos B = (a² + c² – b²)/2 … (ii)

Let us subtract equation (ii) from (i) we get,

ac cos B – bc cos A = (a² + c² – b²)/2 – (b² + c² – a²)/2

= a² – b²

∴ c(a cos B – b cos A) = a² – b²

Thus proved.

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