# prove lhs = rhs   (tan a + sec a – 1)/ (tan a – sec a + 1) = sec a + tan a

Satyajit Samal
9 years ago


Multiplying denominator and numerator bywe get,

Prakash Vaithyanathan
11 Points
7 years ago
if (x / y ) = z to be proved equivalently it is enough to show x = y * z

here y = tan A – sec A + 1  and z = sec A + tan A and x = tan A + sec A – 1

y  * z = tanA – secA + ( sec A + tan A) = sec A + tan A -1

-Prakash Vaithyanathan
Science and Math Tutor
Chennai

Amit Tiwari
35 Points
7 years ago
Amit Tiwari Preparation for IIT form Kanpur. (Kakadev)[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]
Amit Tiwari
35 Points
7 years ago
Amit Tiwari Preparation for IIT from Kanpur (kakadev)[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]