Question

CosA.cosecA - sinA.secA / cosA + sinA = cosecA - secA

Answer 1

(cosA.cosecA-sinA.secA)/(cosA+sinA) = cosecA-secA

we know that ( cosecA = 1/sinA ) and ( secA = 1/cosA). replace cosec and sec with sin and cos and bring the denominator (cosA+sinA) to R.H.S

=> (cosA/sinA)-(sinA/cosA) = [(1/sinA)-(1/cosA)]*(cosA+sinA)

=> (cosA/sinA)-(sinA/cosA) = [(cosA-sinA)/(sinAcosA)]*(cosA+sinA)

=>(cos^2A-sin^2A)/sinAcosA = (cos^2A-sin^2A)/sinAcosA

[ Since (a+b)(a-b) = (a^2 - b^2) ]

=> L.H.S = R.H.S

Hence Proved