Prove the following figure:
sinA-cosA+1/ = 1/
sinA+cosA-1 secA-tanA
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Answer:
(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^2A=sec ^2A
Or 1=sec^2A-tan ^2A=(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.
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