Question
Prove that: $\text{If}\cos\text{A}+\cos\text{B}=\frac{1}{2}\text{ and }\sin\text{A}+\sin\text{B}=\frac{1}{4},$ prove that $\tan\Big(\frac{\text{A+B}}{2}\Big)=\frac{1}{2}.$

Answer

We have, $\text{LHS}=\cos\text{A}+\cos\text{B}=\frac{1}{2}$ $\text{and},\ \sin\text{A}+\sin\text{B}=\frac{1}{4}$ Now, $\frac{\sin\text{A}+\sin\text{B}}{\cos\text{A}+\cos\text{B}}=\frac{\frac{1}{4}}{\frac{1}{2}}$ [On dividing] $\Rightarrow\ \frac{2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}=\frac{1}{2}$ $\Rightarrow\ \frac{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}=\frac{1}{2}$ $\Rightarrow\ \tan\Big(\frac{\text{A}+\text{B}}{2}\Big)=\frac{1}{2}$ $\Rightarrow\ \text{RHS}$ Hence proved.

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