Question

Find the acute angles $A$ and $B$ satisfying
$\cot (A+B)=1,\csc (A-B)=2$

Answer 1

We have,

$\cot (A+B)=1$ and $\csc (A-B)=2$

$\cot (A+B)=\cot {45}^o$

$A+B={45}^o......\left(1\right)$

$\csc (A-B)=2$

$\csc (A-B)=\csc {30}^o$

$A-B={30}^o......\left(2\right)$

On adding equation (1) and (2) to ,we get

$2B={75}^O$

$B=\frac{{75}^o}{2}$

$B=\frac{{75}^o}{2}\times \frac{\pi }{{180}^o}$

$B=\frac{5\pi }{2\times 12}$

$B=\frac{5\pi }{24}$

Put the value of B in equation (1) and we get

$A+B={45}^o$

$A+\frac{{75}^o}{2}={45}^o$

$A={45}^o-\frac{{75}^o}{2}$

$A=\frac{{90}^o-{75}^o}{2}$

$A=\frac{{15}^o}{2}$

$A=\frac{{15}^o}{2}\times \frac{\pi }{{180}^o}$

$A=\frac{\pi }{24}$

Hence, this is the answer.