prove that inA-cosA+1÷sinA+cosA-1=1÷secA-tanA using identity sec2A = 1+tan2A

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Solution

(sin A-cos A+1)/(sin A+cosA-1)=1/(sec A-tan A)

L.H.S. divide above and below by cos A

=(tan A-1+secA)/(tan A+1-sec A)

=(tan A-1+secA)/(1-sec A+tan A)

We know that 1+tan^2(A)=sec ^2(A)

Or put in deminator 1=sec^2(A)-tan ^2(A)=(sec A+tan A)(secA-tanA)

=(sec A+tan A-1)/[(sec A+tan A)(sec A-tan A)-(sec A-tan A)]

=(sec A+tan A-1)/(sec A-tan A)(sec A+tan A-1)

= 1/(sec A-tan A) , proved.
LHS =RHS

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