I need help simplifying this trig expression. I started by trying to find a common denom. btwn the two fractions

Originally Posted by hunnyelle

Simplify [(cosxcotx)/(secx+tanx)] + (sinx)/(secx-tanx)]

I need help simplifying this trig expression. I started by trying to find a common denom. btwn the two fractions

note that your common denominator, $\displaystyle \sec^2{x} - \tan^2{x}$ equals 1


btw ... do not double post.

@skeeter: Yes , but then I'm left w/ (cosxcotx(secx-tanx) + sinx(sexc + tanx). I tried putting all of tht in terms of sin & cos & eventually reached the point [(cos^2x(cosx-sinx)+ sin^2x(1+sinx)]/(sinxcosx)]

Yes, when you double post, and your duplicate post gets deleted, then the help in the duplicate gets deleted as well, wasting time and effort of those trying to help.

$\displaystyle \cos{x}\cot{x}(\sec{x}-\tan{x}) + \sin{x}(\sec{x}+\tan{x})$

$\displaystyle \cos{x}\cot{x}\sec{x} - \cos{x}\cot{x}\tan{x} + \sin{x}\sec{x} + \sin{x}\tan{x}$

$\displaystyle \cot{x} - \cos{x} + \tan{x} + \sin{x}\tan{x}$

$\displaystyle \cot{x}(1 - \sin{x}) + \tan{x}(1 + \sin{x})$

I don't think changing every term to sines and cosines will do much good from this point ... do you know what it is supposed to simplify to?

The answer is cotx -cosx + tanx +sinxtanx, & I've figured out how to reach that answer!