# [SOLVED] Prove the identity, Stuck?

• Aug 7th 2009, 06:38 AM
mooniq
[SOLVED] Prove the identity, Stuck?
Hi, I tried doing this equation over and over, but I can't seem to get it! I've had sleepless nights for the past week...lol. So, please help me and put me out of my misery, I beg you...

I need to prove that:
$\displaystyle \frac {1+sinA}{cosA} + \frac{cosA}{1+sinA} = 2secA$

this is what I came up with so far on the LHS, using $\displaystyle 1=sinA+cosA$:

$\displaystyle = \frac {(sinA+cosA)+sinA}{cosA} + \frac {cosA}{(sinA+cosA)+sinA}$

$\displaystyle = \frac {sinA+cosA+sinA}{cosA} + \frac{cosA}{sinA+cosA+sinA}$

$\displaystyle = \frac {(1-cosA)+cosA+(1-cosA)}{cosA} + \frac{cosA}{(1-cosA)+cosA+(1-cosA)}$

$\displaystyle = \frac {2-cosA}{cosA} + \frac{cosA}{2-cosA}$

And that's where I got stuck. I don't even think my working out is right...

PS. I tried using Wolfram|Alpha to solve it but it just tells me that the equation is true and gives me alternate forms, it doesn't show me steps which I need in order to fully understand, ya know?
• Aug 7th 2009, 06:45 AM
Failure
Quote:

Originally Posted by mooniq
Hi, I tried doing this equation over and over, but I can't seem to get it! I've had sleepless nights for the past week...lol. So, please help me and put me out of my misery, I beg you...

I need to prove that:
$\displaystyle \frac {1+sinA}{cosA} + \frac{cosA}{1+sinA} = 2secA$

this is what I came up with so far on the LHS, using $\displaystyle 1=sinA+cosA$:

$\displaystyle = \frac {(sinA+cosA)+sinA}{cosA} + \frac {cosA}{(sinA+cosA)+sinA}$

$\displaystyle = \frac {sinA+cosA+sinA}{cosA} + \frac{cosA}{sinA+cosA+sinA}$

$\displaystyle = \frac {(1-cosA)+cosA+(1-cosA)}{cosA} + \frac{cosA}{(1-cosA)+cosA+(1-cosA)}$

$\displaystyle = \frac {2-cosA}{cosA} + \frac{cosA}{2-cosA}$

And that's where I got stuck. I don't even think my working out is right...

PS. I tried using Wolfram|Alpha to solve it but it just tells me that the equation is true and gives me alternate forms, it doesn't show me steps which I need in order to fully understand, ya know?

$\displaystyle \begin{array}{lcl} \displaystyle \frac {1+\sin A}{\cos A} + \frac{\cos A}{1+\sin A} &=& \displaystyle \frac{(1+\sin A)^2+\cos^2 A}{\cos A\cdot (1+\sin A)}\\ &=& \displaystyle \frac{2(1+\sin A)}{\cos A\cdot (1+\sin A)}\\ &=& \displaystyle \frac{2}{\cos A}\\ &=& 2\sec A \end{array}$
• Aug 7th 2009, 06:46 AM
malaygoel
you are using the wrong identity

Correct:
$\displaystyle 1-sin^2A+cos^2A$

$\displaystyle 1=sinA+cosA$