Question

If $\tan \left(A\right)-\tan \left(B\right)=x$ and $\cot \left(B\right)-\cot \left(A\right)=y$, then $\cot \left(A-B\right)=?$

• A. $\frac{1}{x}+y$
• B. $\frac{1}{xy}$
• C. $\frac{1}{x}-\frac{1}{y}$
• D. $\frac{1}{x}+\frac{1}{y}$

Answer 1

The correct option is D

$\frac{1}{x}+\frac{1}{y}$

Explanation for the correct option

Given $\tan \left(A\right)-\tan \left(B\right)=x$ and $\cot \left(B\right)-\cot \left(A\right)=y$

$\begin{array}{ccccc}\therefore & \cot \left(B\right)-\cot \left(A\right)& =& y& \\ \Rightarrow & \frac{1}{\tan \left(B\right)}-\frac{1}{\tan \left(A\right)}& =& y& \left[\because \cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}\right]\\ \Rightarrow & \frac{\tan \left(A\right)-\tan \left(B\right)}{\tan \left(B\right)\tan \left(A\right)}& =& y& \\ \Rightarrow & \frac{x}{\tan \left(B\right)\tan \left(A\right)}& =& y& \left[\because \tan \left(A\right)-\tan \left(B\right)=x\right]\\ \Rightarrow & \tan \left(B\right)\tan \left(A\right)& =& \frac{x}{y}& \\ & & & & \\ \therefore & \cot \left(A-B\right)& =& \frac{1}{\tan \left(A-B\right)}& \left[\because \cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}\right]\\ & & =& \frac{1}{{\displaystyle \frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}}}& \left[\because \tan \left(\alpha -\beta \right)=\frac{\tan \left(\alpha \right)-\tan \left(\beta \right)}{1+\tan \left(\alpha \right)\tan \left(\beta \right)}\right]\\ & & =& \frac{1+\tan \left(A\right)\tan \left(B\right)}{\tan \left(A\right)-\tan \left(B\right)}& \\ & & =& \frac{1+{\displaystyle \frac{x}{y}}}{x}& \\ & & =& \frac{x+y}{xy}& \\ \Rightarrow & \cot \left(A-B\right)& =& \frac{1}{x}+\frac{1}{y}& \end{array}$

Hence the correct option is option(D) i.e. $\frac{1}{x}+\frac{1}{y}$