3
$\begingroup$

I have this problem: Prove that $\csc (x) +\cot( x)=\dfrac{\sin (x)}{1-\cos(x)}$

From LHS I tried using $\sin^2x+\cos^2x = 1$ and ended up nowhere. I tried rearranging RHS but ended up with $\dfrac{1+\cos (x)}{\sin (x)}$. I'm really stuck with this one. Any suggestions?

$\endgroup$
  • $\begingroup$ I reformatted your question. Please check that I didn't alter the meaning of what you typed. $\endgroup$ – Git Gud Nov 27 '13 at 13:39
2
$\begingroup$

You have $\dfrac{\sin(x)}{1-\cos(x)}$ Mutiply numerator and denominator by $1+\cos x$ to get $\dfrac{\sin x(1+\cos x)}{1-\cos^2(x)}$

Now $1-\cos^2(x)= \sin^2(x)$. Therfore we are left with $\dfrac{1+\cos(x)}{\sin(x)}$ which is $\csc(x) +\cot(x)$:

$\csc(x) + \cot(x)$ = $\dfrac{1}{\sin(x)} + \dfrac{\cos(x)}{\sin(x)}$ = $\dfrac{1 + \cos(x)}{\sin(x)}$

$\endgroup$
  • $\begingroup$ Thank you! But how did you figure to multiply numerator and denominator by 1 + cosx? $\endgroup$ – user111835 Nov 27 '13 at 13:47
  • $\begingroup$ It is just practice and experience. These methods will click fast if you get enough of practice. $\endgroup$ – Apurv Nov 27 '13 at 13:48
  • 1
    $\begingroup$ @user111835 That's call multiplying by its conjugate and because you multiply a set of terms with only the sign in the middle being different, you'll end up with a difference of squares. The difference of squares you would end up with in this case would be a trig identity. $\endgroup$ – Michael Yaworski Jan 18 '14 at 6:27
1
$\begingroup$

Or from the left : csc + cot = (1 + cos)/sin then multiply top and bottom by (1 - cos) and simplify using diff of squares and sin^2 + cos^2 = 1 identity.

$\endgroup$
1
$\begingroup$

We have $$\cos^2x+\sin^2x=1$$

$$\iff \sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$$

$$\implies \frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}=\frac1{\sin x}+\frac{\cos x}{\sin x}=\cdots$$

$\endgroup$

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.