Trigonometry> prove lhs = rhs (tan a + sec a – 1)/ (tan...

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

Prateek Agrawal , 11 Years ago

Grade 12

4 Answers

Satyajit Samal

$\frac{\tan a + \sec a -1}{\tan a -\sec a +1}= \frac{\sin a +1 - \cos a}{\sin a -1 + \cos a}\\$
$=\frac{2 \sin {a/2} \cos {a/2}+2 \sin ^{2}{a/2}}{2 \sin {a/2} \cos {a/2}-2 \sin ^{2}{a/2}} \\ = \frac{\cos {a/2}+ \sin {a/2}}{\cos {a/2}-\sin {a/2}}$
Multiplying denominator and numerator by $\cos {a/2}+ \sin {a/2}$ we get,
$\frac{1+ \sin a}{\cos a} \\ = \sec a + \tan a$

Approved

Last Activity: 11 Years ago

Prakash Vaithyanathan

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 = tan²A – sec²A + ( sec A + tan A) = sec A + tan A -1

-Prakash Vaithyanathan

Science and Math Tutor

Chennai

Last Activity: 9 Years ago

Amit Tiwari

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]

Last Activity: 9 Years ago

Amit Tiwari

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]

Last Activity: 9 Years ago