If \(\tan A-\tan B=x\) and \(\cot A-\cot B=y\), then \(\cot (A-B)=\)
If \(\tan A-\tan B=x\) and \(\cot A-\cot B=y\), then \(\cot (A-B)=\)
- \(\frac{x y}{x+y}\)
- \(\frac{x y}{x-y}\)
- \(\frac{x-y}{x y}\)
- \(\frac{y-x}{x y}\)
Solution
Since, \(\tan A-\tan B=x\)
\(\Rightarrow \frac{1}{\cot A}-\frac{1}{\cot B}=x \Rightarrow \frac{\cot B-\cot A}{\cot A \cot B}=x\)
\(\because \quad \cot A-\cot B=y\) (given)
So, \(\quad \cot A \cot B=-\frac{y}{x}\)
\(\because \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}=\frac{-\frac{y}{x}+1}{-y}=\frac{y-x}{x y}\)
Hence, option (d) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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