Question

if (sinA+cosA)=m and (secA+cosecA)=n , then prove that n(m.m -1)=2m

Answer 1

n(m*m-1)=2m
((1/sinA)+(1/cosA))(sin^2A+cos^2+2sinAcosA-1)=2(sinA+cosA)

((1/sinA)+(1/cosA)(2sinAcosA)=sinA+cosA

(sinA+cosA)*sinAcosA/sinAcosA=sinA+cosA

sinA+cosA=sinA+cosA

0=0