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Prove that: $$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$

My Attempt:

$$L.H.S=\csc 50° + \cot 100°=\frac {1}{\sin 50°} + \frac {\cos 100°}{\sin 100°}$$ $$=\frac {1}{\sin 50°} + \frac {\cos 100°}{2\sin 50°\cos 50°}$$ $$=\frac {2\cos 50° + \cos 100°}{2\sin 50°\cos 50°}$$.

Now, what should I do?

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  • $\begingroup$ i think your statement is wrong, have you try it with a calculator? $\endgroup$ Dec 30, 2016 at 15:51
  • $\begingroup$ Simple consequence of the doubling formulas $\cos 2x=2\cos^2x-1$ and $\sin 2x=2\sin x\cos x$, hence for every $a$ such that one does not divide by $0$, $$S=\frac{1}{\sin 2a}+\frac{\cos 4a}{\sin 4a}+\frac{1}{\sin 4a}=\frac{1}{\sin 2a}+\frac{1+\cos 4a}{\sin 4a}$$ is also $$S=\frac{1}{\sin 2a}+\frac{2\cos^2 2a}{2\sin 2a\cos 2a}=\frac{1}{\sin 2a}+\frac{\cos 2a}{\sin 2a}=\frac{1+\cos 2a}{\sin 2a}$$ which is also $$S=\frac{2\cos^2a}{2\sin a\cos a}=\frac{\cos a}{\sin a}$$ Now, use this for $$a=25°$$ $\endgroup$
    – Did
    Dec 30, 2016 at 15:52
  • $\begingroup$ yes it is true, it was a typo of mine, thank you $\endgroup$ Dec 30, 2016 at 15:54
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    $\begingroup$ @Dr.SonnhardGraubner "Typo", meaning that you mistyped when you entered the formula into your CAS, yes we know (and nobody is interested, as you have repeatedly been told). $\endgroup$
    – Did
    Dec 30, 2016 at 15:55
  • $\begingroup$ @Dr. Sonnhard Graubner, with using a calculator I got both R.H.S=L.HS.=1.129080309. $\endgroup$
    – pi-π
    Dec 30, 2016 at 15:55

2 Answers 2

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This would work better may be : $$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$ $$\csc 50° + \cot 100° +\csc 100° = \cot 25° $$ $$\frac{1}{\sin 50°} + \frac{\cos 100°}{\sin 100°} +\frac{1}{\sin 100°} = \cot 25° $$ $$\frac{1}{\sin 50°} + \frac{\cos 100° + 1}{\sin 100°} = \cot 25° $$ $$\frac{1}{\sin 50°} + \frac{2\cos^2 50° }{2\sin 50°\cos 50°} = \cot 25° $$ $$\frac{1}{\sin 50°} + \frac{\cos 50° }{\sin 50°} = \cot 25° $$ $$\frac{1+\cos 50° }{\sin 50°} = \cot 25° $$ $$\frac{2\cos^2 25° }{2\sin 25°\cos 25°} = \cot 25° $$

Done right?

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  • $\begingroup$ Hello win vineeeth, sir. very wonderful solution. Do I get your email address or any thing else so that I can contact you out of MSE as well? $\endgroup$
    – pi-π
    Dec 30, 2016 at 16:12
  • $\begingroup$ @user354073 I would prefer not. You can use MSE itself to ask me anything. I also want to mention that I am not as good as some of the people here on MSE. Anyway, Glad you liked my solution. Upvote and accept it if worth. Thanks. $\endgroup$ Dec 30, 2016 at 16:15
  • $\begingroup$ But I have some queries that I want to ask you personally, and get some help.? $\endgroup$
    – pi-π
    Dec 30, 2016 at 16:22
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Hint:

$\csc2x+\cot2x=\dfrac{1+\cos2x}{\sin2x}=\dfrac{2\cos^2x}{2\cos x\sin x}=\cot x$

Set $x=50^\circ,100^\circ$

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  • $\begingroup$ @ lab bhattacharjee, I request you, please once have a look at this question, math.stackexchange.com/questions/2106883/graph-third-quartle/…. The answer I got here doesn't match with the one provided in my book. The boys says that the answers are $20-30$ and $30$. $\endgroup$
    – pi-π
    Jan 28, 2017 at 9:33
  • $\begingroup$ @NeWtoN, Not sure if I have understood u correctly. $\endgroup$ Jan 28, 2017 at 9:34
  • $\begingroup$ Hello sir, did you see the question. @lab bhattacharjee $\endgroup$
    – pi-π
    Jan 28, 2017 at 9:41

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