# secx + tanx = 1/(secx-tanx)

• January 20th 2013, 02:12 AM
Furyan
Solved :secx + tanx = 1/(secx-tanx)

$\sec x + \tan x \equiv \dfrac{1}{\sec x -\tan x}$

I got as far as this starting with the R.H.S, but don't know if it's the right approach or how to proceed from here.

$R.H.S = \dfrac{\cos x}{1 - \sin x}$

Any help would be much appreciated, thank you.

Now solved. I multiplied the denominator and numerator by $1 + \sin x$, then it was easy.
• January 20th 2013, 02:45 AM
ibdutt
Re: secx + tanx = 1/(secx-tanx)
You are correct, now just write it as 1/[(1-sinx)/cosx] and proceed further.
Alternatively multiply and divide the numerator and denominator by ( secx-tanx) the numerator will reduce to 1 and you will get the RHS.
• January 20th 2013, 02:57 AM
Furyan
Re: secx + tanx = 1/(secx-tanx)
Thank you very much for your reply, sorry I did not say in my original post, but I actually started with the R.H.S. I've now edited my post accordingly thank you.