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americalvr · Feb 26, 2008, 05:25 PM

somebody pleease help me. :confused: I've tried figuring out this problem and can't seem to get it. :mad: you have to verify the trig identity

the problem is
(cotx/cosx)+(secx/cotx)=sec2xcscx

pleeease help me.

rockerchick_682 · Feb 26, 2008, 05:36 PM

Uh it's been a while. But have you tried converting them all to cosx and sinx? Isn't cotx cosx over sinx? And secx is 1/cosx? Well I'd try that. Keep trying to substitute all the rules in

galactus · Feb 27, 2008, 07:51 AM

You have some bad notation. You have $sec(2x)$ when you mean $sec^{2}(x)$ . Please use the ^ for powers. Better yet, if you plan on posting to any extent, try learning a little LaTex. If you wish, you can click on 'quote' at the bottom of my post to see the code I used to display it.

$\frac{cot(x)}{cos(x)}+\frac{sec(x)}{cot(x)}=sec^{2 }(x)csc(x)$

The previous advise was good. Convert the left side to sin and cos.

$\frac{\frac{cos(x)}{sin(x)}}{cos(x)}+\frac{\frac{1 }{cos(x)}}{\frac{cos(x)}{sin(x)}}$

$\frac{1}{sin(x)}+\frac{sin(x)}{cos^{2}(x)}$

Now, can you continue? A little cross multiplying and you should see it fall together.. Write back if you remain stuck. Just remember your identities. $sin^{2}(x)+cos^{2}(x)=1$ may prove handy.