Question

If $x=\tan A-\tan B,y=\cot B-\cot A$ then $\frac{1}{x}+\frac{1}{y}=$

• A. $\cot (A-B)$
• B. $\cot (B-A)$
• C. $\tan (A-B)$
• D. $\tan (B-A)$

Answer 1

Answer: (B)

$\cot (B-A)$

$\tan A-\tan B=x,\cot B-\cot A=y$

We have $\tan A-\tan \beta =x$

$=\frac{\sin (A-B)}{\cos A\cos B}$

$\cot B-\cot A=\frac{\sin (A-B)}{\sin A\sin B}=y$

$\frac{1}{x}+\frac{1}{y}=\frac{\cos A\cos B}{\sin (A-B)}+\frac{\sin A\sin B}{\sin (A-B)}$

$=\frac{\cos A\cos B+\sin A\sin B}{\sin (A-B)}=\frac{\cos (A-B)}{\sin (A-B)}$

$\frac{1}{x}+\frac{1}{y}=\cot (A-B)$.