Question

If x = cosec A + cos A and y = cosec A – cos A then prove that `(2/(x + y))^2 + ((x - y)/2)^2 = 1`.

Answer 1

LHS = `(2/(x+y))^2 + ((x-y)/2)^2 = 1`

= `[2/((cosec A + cos A)+(cosec A - cos A))]^2 + [((cosecA+cosA)-(cosecA-cosA))/2]^2 - 1`

= `[2/(cosecA + cosA + cosecA-cosA)]^2 + [(cosecA +cosA-cosecA+cosA)/2]^2-1`

= `[2/(2cosecA)]^2 + [(2 cosA)/2]^2-1`

= `[1/(cosecA)]^2 + [cosA]^2-1`

= `[sinA]^2 + [cosA]^2-1`

= `sin^2 A + cos^2 A-1`

= 1 – 1

= 0

= RHS