Question

(tanTheta+secTheta-1)/(tanTheta-secTheta+1) = (1+sinTheta)/(cosTheta) Prove both expressions are equal.

Answer 1

\[\frac{ \tan(\theta)+\sec(\theta)-1 }{ \tan(\theta)-\sec(\theta) +1 }= \frac{ 1+\sin(\theta) }{ \cos(\theta) }\]

Answer 2

thats how the equation looks like right?

Answer 3

yes thats exactly how it looks

Answer 4

I've got up to the point where you can rationalize, but after that I get stuck.

Answer 5

so substitute tan(theta)= sin(theta)/cos(theta)

Answer 6

sec(theta)= 1/cos(theta)

Answer 7

I did, I nearly finished the problem its just that I don't know what to rationalize

Answer 8

My near finished problem looks like this : sintheta+1-costheta/sintheta-1+costheta

Answer 9

Do you rationalize the top or the bottom to make the end result 1+sintheta/costheta?

Answer 10

wait , lemme work it out on paper..

Answer 11

thanks

Answer 12

this is how far i got:\[\frac{ \frac{ \sin(\theta)+1-\cos(\theta) }{ \cos(\theta)} }{ \frac{ \sin(\theta)-1+\cos(\theta) }{ \cos(\theta) } }= \frac{ 1+\sin(\theta) }{ \cos(\theta) }\]

Answer 13

thats fine thank you for attempting to solve my problem

Answer 14

@sauravshakya

Answer 15

\[\frac{\tan x+\sec x-1}{\tan x-\sec x+1}\]multiply num and denum by \(\cos x\) u will getv\[\frac{\sin x+1-\cos x}{\sin x-1+\cos x}=\frac{\sin x+1-\cos x}{\sin x-(1-\cos x)}\]now use these identities\[\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\]\[1-\cos x=2 \sin^2 \frac{x}{2}\]does that help...?!!!

Answer 16

thanks so much.

Answer 17

yw :)