Question

if log tan A+log tan B=0 then find log sin(A+B) where A and B are acute angles

Answer 1

$\textbf{Given:}$

$log\,tanA+log\,tanB=0$

$\text{Using product rule of logarithm}$

$log\,(tanA\,tanB)=0$

$\implies\;tanA\,tanB=1$

$\implies\;1-tanA\,tanB=0$

$\text{Now,}$

$tan(A+B)=\displaystyle\frac{tanA+tanB}{1-tanA\,tanB}$

$tan(A+B)=\displaystyle\frac{tanA+tanB}{0}$

$tan(A+B)=\infty$

$\implies\;A+B=90^{\circ}$

$\text{consider}$

$log\,sin(A+B)$

$=log\,sin90^{\circ}$

$=log\,1$

$=0$

$\therefore\boxed{\bf\,log\,sin(A+B)=0}$