Question

If $\tan A-\tan B=x,\cot B-\cot A=y$ and$\left(\sec \right(A-B)-1)\left(\sec \right(A-B)+1)=\frac{x^a{y}^b}{(x+y{)}^c},$ then find the value of $a+b+c$.

Answer 1

If $\tan A-\tan B=x,\cot B-\cot A=y..................$
$\left(\sec \right(A-B)-1)\left(\sec \right(A-B)+1)$
$={\sec }^2(A-B)-1={\tan }^2(A-B)=(\frac{\tan A-\tan B}{1+\tan A\tan B}{)}^2.....\left(i\right)$
$y=\cot B-\cot A=\frac{\tan A-\tan B}{\tan A\tan B}=\frac{x}{\tan A\tan B}$
$\therefore \tan A\tan B=\frac{x}{y}.............\left(ii\right)$
from (i) and (ii), we get
$\left(\sec \right(A-B)-1)\left(\sec \right(A-B)+1)=(\frac{x}{1+\left(x/y\right)}{)}^2$
$=\frac{x^2{y}^2}{(x+y{)}^2}$
$\therefore a=b=c=2$
$\therefore a+b+c=6$