Question

Find the acute angles A and B satisfying

(a) $cot(A+B)=1,cosec(A-B)=2$

(b) $secAcotB-secA-2cotB+2=0$

Answer 1

(a)

$\begin{array}{c}\cot \left(A+B\right)=1,\cos ec\left(A-B\right)=2\\ A+B=\frac{\pi }{4},A-B=\frac{\pi }{6}\end{array}$

Adding two equations,

$\begin{array}{c}A+B+A-B=\frac{\pi }{4}+\frac{\pi }{6}\\ 2A=\frac{10\pi }{24}\\ A=\frac{5\pi }{24}\\ and\\ B=\frac{\pi }{4}-\frac{5\pi }{24}\\ =\frac{\pi }{24}\end{array}$

(b)

$\begin{array}{c}\sec A\cot B-\sec A-2\cot B+2=0\\ \sec A\left(\cot B-1\right)-2\left(\cot B-1\right)=0\\ \left(\sec A-2\right)\left(\cot B-1\right)=0\\ \sec A-2=0,\cot B-1=0\\ \sec A=2,\cot B=1\\ A=\frac{\pi }{3},B=\frac{\pi }{4}\end{array}$