Question

If $\tan A-\tan B=x$, $\cot B-\cot A=y$, prove that $\cot (A-B)=\frac{1}{x}+\frac{1}{y}$.

Answer 1

$\begin{array}{c}\tan A-\tan B=x.\cot B-\cot A=y\\ \cot +\left(A-B\right)=\frac{1}{x}+\frac{1}{y}\\ weknow\\ \cot \left(A-B\right)=\frac{\cot A-\cot B+1}{\left(\cot B-\cot A\right)}\to \left(i\right)\\ \tan \frac{\tan A-\tan B}{1+5\tan A.\tan B\to \left(ii\right)}\\ fromequation\left(i\right)\\ \Rightarrow \cot \left(A-B\right)=\frac{\cot A.\cot B}{\cot B-\cot A}+\frac{1}{\cot B-\cot A}\\ \Rightarrow \cot \left(A-B\right)=\frac{\cot A.\cot B}{\frac{1}{\tan B}-\frac{1}{\tan A}}+\frac{1}{y}=\frac{\cot A.\tan A.\cot B.\tan B}{\tan A-\tan B}+\frac{1}{y}-\frac{1}{x}+\frac{1}{y}\\ S.P=2100rupees.\end{array}$