Prove that: + - = ( A-B 2 ) ( A-B 2 ) - Vidyadip

Question

Prove that: $\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$

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Answer

We have, $\text{LHS}=\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}$ $=\ \frac{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-2\sin\Big(\frac{\text{B}+\text{A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}$ $=\ \frac{-\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}$ $=\ \frac{-\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}$ $= \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $=\ \text{RHS}$ $\therefore\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}=\cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big).$ Hence proved.

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