Question

Find the acute angles A and B satisfying $\sec A\cot B-\sec A-2\cot B+2=0$.

Answer 1

Q. Find the acute angles A & B

satisfying

$secAcotB-secA-2cotB+2=0$

Sol${}^n\Rightarrow$ We are given,

$secAcotB-secA-2cosB+2=0$

$secAcotB-secA=2cosB-2$

$secA(cotB-1)=2(cosB-1)$

$secA(cotB-1)-2(cosB-1)=0$

$(secA-2)(cotB-1)=0$

So, $secA=2$ or $cotB=1$

now, $A,BE(0,\frac{\pi }{2})$

$A=se{c}^{-1}\left(2\right)$ & $B=co{t}^{-1}\left(1\right)$

$=\frac{\pi }{3}$ $=\frac{\pi }{4}$

we get $A=\frac{\pi }{3}\&B=\frac{\pi }{4}$