If two acute angles $A$ and $B$ are such that $A \neq B$ and $\frac{x}{y}=\frac{\cos A}{\cos B}$, then…

If two acute angles $A$ and $B$ are such that $A \neq B$ and $\frac{x}{y}=\frac{\cos A}{\cos B}$, then $\frac{x \tan A-y \tan B}{x+y}=$
  1. $\tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$
  2. $\tan \left(\frac{B-A}{2}\right)$
  3. $\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)$
  4. $\cot \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)$

Solution

Given : $\frac{x}{y}=\frac{\cos A}{\cos B}$ $\frac{x \tan A-y \tan B}{x+y}=\frac{\frac{x}{y} \tan A-\tan B}{\frac{x}{y}+1}$ $=\frac{\frac{\cos A}{\cos B} \cdot \tan A-\tan B}{\frac{\cos A}{\cos B}+1}$ $\begin{aligned} & =\frac{\cos A \cdot \tan A-\tan B \cos B}{\cos A+\cos B} \\ & =\frac{\sin A-\sin B}{\cos A+\cos B}\end{aligned}$ $=\frac{2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)}{2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}$ $\Rightarrow \frac{x \tan A-y \tan B}{x+y}=\tan \left(\frac{A-B}{2}\right)$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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