prove that .tanA + secA -1 / tanA -secA + 1 = 1 +sinA / cosA

Hello Pranav, your question is unclear that which terms are together or which terms are present in the numerator and denominator... I request you to resend the question by using proper brackets to separate the terms....

[tanA + secA - 1] / [tanA-secA+1]= [ tanA + secA - (sec2A-tan2A) ] / [tanA-secA+1] [Using: sec2A-tan2A=1]= [ (tanA+secA) - (secA+tanA)(secA-tanA) ] / [tanA-secA+1] [Using: a2-b2=(a+b)(a-b)]= [ (secA+tanA)*(1- (secA-tanA) ) ] / [tanA-secA+1] [Taking (secA+tanA) as common]= [ (secA+tanA)*(1-secA+tanA) ] / [tanA-secA+1]= [ (secA+tanA)*(tanA-secA+1) ] / [tanA-secA+1]= secA + tanA= (1/cosA) + (sinA/cosA)= (1+sinA)/cosA [Taking LCM]