Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec² A = 1 cot² A.

Answer 1

`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA`

Using the identity cosec²A = 1 + cot²A,

L.H.S = `(cos A-sinA+1)/(cosA+sinA-1)`

= `(cosA/sinA-sinA/sinA+1/sinA)/(cosA/sinA+sinA/sinA+1/sinA)`

= `(cotA-1+cosec  A)/(cotA+1-cosec  A)`

= `({(cotA)-(1-cosec  A)}{(cotA)-(1-cosec  A)})/({(cotA)+(1-cosec  A)}{(cotA)-(1-cosec  A)})`

= `(cot A - 1 + cosecA)^2/((cotA)^2-(1-cosecA)^2)`

= `(cot^2A+1+cosec^2A-2cotA-2cosec  A+2cotAcosec  A)/(cot^2A-(1+cosec^2  A-2cosec  A))`

= `(2cosec^2  A+2cotAcosec  A-2cotA-2cosec  A)/(cot^2A-1-1cosec^2  A+2cosec  A)`

= `(2cosec  A(cosecA+cotA)-2(cotA+cosec  A))/(cot^2A-cosec^2A-1+2cosec  A)`

= `((cosec  A+cotA)(2cosec  A-2))/(-1-1+2cosec  A)`

= `((cosec  A+cotA)(2cosec  A-2))/(2cosec  A-2)`

= cosec A + cot A

= R.H.S