For any Δ ABC show that-

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

The key point to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

The idea of sine formula in ΔABC

•

As we have to prove:

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

As LHS contain (c² – a² + b²), (a² – b² + c²)and (b² – c² + a²),which shows resemblance with cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ 2bc cos A =

Multiplying with tan A both sides to get the form desired in proof

2bc cos A tan A =

2bc sin A = …..eqn 1

Cos C =

⇒ 2ab cos C =

Multiplying with tan C both sides to get the form desired in proof

2ab cos C tan C = tan C

2ab sin C = tan C ..…eqn 2

And, Cos B =

⇒ 2ac cos B =

Multiplying with tan B both sides to get the form desired in proof

2ac cos B tan B =

2ac sin B = ……eqn 3

As we are observing that sin terms are being involved so let’s try to use sine formula.

From sine formula we have,

⇒

Multiplying abc to each fraction:-

⇒ bc sin A = ac sin B = ab sin C

⇒ 2bc sin A = 2ac sin B = 2ab sin C

∴ From eqn 1, 2 and 3 we have:

= tan C

Hence, proved.