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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?

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I reformatted your question. Please check that I didn't alter the meaning of what you typed. –  Git Gud Nov 27 '13 at 13:39
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3 Answers 3

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 $\csc(x) +\cot(x)$:

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

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Thank you! But how did you figure to multiply numerator and denominator by 1 + cosx? –  user111835 Nov 27 '13 at 13:47
    
It is just practice and experience. These methods will click fast if you get enough of practice. –  Apurv Nov 27 '13 at 13:48
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@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. –  mike yaworski Jan 18 at 6:27
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We have $$\cos^2x+\sin^2x=1$$

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

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

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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.

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