Question

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
(v) `(cosA-sinA+1)/(cosA+sinA-1)=cosec A+cotA`

Answer 1

`L.H.S. = (cosA -sinA+1)/(cosA+sinA-1)`
Multiplying Numerator and denominator by `cosA+sinA+1`
`=(cosA -sinA+1)/(cosA+sinA-1)xx(cosA+sinA+1)/(cosA+sinA+1)`
`=((1+cosA)^2-(sinA)^2)/((cosA+sinA)^2-(1)^2)`
`=(1+cos^2a+2cosA-sin^2A)/(cos^2A+sin^2A+2sinAcosA -1)`
As, `cos^2A+sin^2A = 1`,our equation becomes,
`=(cos^2A+sin^2A+cos^2a+2cosA-sin^2A)/(1+2sinAcosA -1)`
`(2cos^2A+2cosA)/(2sinAcosA) =(1+cosA)/sinA = cosecA+cotA=R.H.S.`