If $\mathrm{A}\gt\mathrm{B}$ and $\tan \mathrm{A}-\tan \mathrm{B}=x$ and $\cot \mathrm{B}-\cot \mathrm{A}=y$…
If $\mathrm{A}\gt\mathrm{B}$ and $\tan \mathrm{A}-\tan \mathrm{B}=x$ and $\cot \mathrm{B}-\cot \mathrm{A}=y$, then $\cot (\mathrm{A}-\mathrm{B})=$
- $\frac{1}{y}-\frac{1}{x}$
- $\frac{1}{x}-\frac{1}{y}$
- $\frac{1}{x}+\frac{1}{y}$
- $\frac{x y}{x-y}$
Solution
Given, $\tan \mathrm{A}-\tan \mathrm{B}=x$
$\cot \mathrm{B}-\cot \mathrm{A}=y$
$\begin{aligned}
& \Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y \\
& \Rightarrow \frac{\tan A-\tan B}{\tan A \cdot \tan B}=y \\
& \Rightarrow \tan A \cdot \tan B=\frac{x}{y}...(i)
\end{aligned}$
$\begin{aligned}
& \text {Now, } \cot (A-B) =\frac{1}{\tan (A-B)} \\
& =\frac{1+\tan A \cdot \tan B}{\tan A-\tan B} \\
& =\frac{1+\frac{x}{y}}{x} \\
& =\frac{y+x}{x y} \\
& =\frac{1}{x}+\frac{1}{y}
\end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 2)
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