# Prove the following identity, where the angle involved is acute angle for which the expressions are defined. (cosA-sinA+1)/cosA+sinA-1)=cosecA+cotA.

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Prove the following identity, where the angle involved is acute angle for which the expressions are defined.

(cosA-sinA+1)/(cosA+sinA-1)=cosecA+cotA.

using the identity cosec2A=1+cot2A.

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by (15 points)
cosA-sinA+1/cosA +sinA-1 = cosecA+cotA

Use the formula cot^2A-cosec^2A

LHS-   cosA-sinA+1/cosA+sinA-1

divide by sinA on both numerator and denominator

cosA/sinA-sinA/sinA+1/sinA /cosA/sinA-sinA/sinA+1/sinA

cotA-1+cosecA/cotA+1-cosecA

at the place of 1 put above formula

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

now put it in a^2-b^2 =(a+b)(a-b)

cotA+cosecA+(cotA-cosecA)(cotA+cosecA)/cotA-cosecA+1

now common taking from numerator

cotA+cosecA(1+cotA-cosecA)/cotA-cosecA+1

Hence it prove cotA+cosecA