0
$\begingroup$

I got an urge for some trigonometry so I dug up my old books and got to it! I also have a lot of my old answers as well but there was one I can't figure out how I was thinking!

I had to simpify this expression:

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

I can't figure out I got from the first to second step.. Any suggestions? Thank you! :)

$\endgroup$
1
  • 1
    $\begingroup$ $\displaystyle a+{b\over c}={a\over 1}+{b\over c}={a\cdot c\over c}+{b\over c}={ac+b\over c} $. $\endgroup$ – David Mitra Jun 7 '15 at 10:15
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.