Prove that ![]()
(sin A – cos A + 1)/(sin A + cos A – 1) = 1/(sec A – tan A)
L.H.S. divide numerator and denominator by cos A
= (tan A – 1 + sec A)/ (tan A + 1 – sec A)
= (tan A – 1 + sec A)/(1 – sec A + tan A)
We know that 1 + tan2 A = sec2A
Or 1 = sec2 A – tan2 A = (sec A + tan A)(sec A – tan A)
= (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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