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

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?

• i think your statement is wrong, have you try it with a calculator? Dec 30, 2016 at 15:51
• 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 $$a=25°$$
– Did
Dec 30, 2016 at 15:52
• yes it is true, it was a typo of mine, thank you Dec 30, 2016 at 15:54
• @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).
– Did
Dec 30, 2016 at 15:55
• @Dr. Sonnhard Graubner, with using a calculator I got both R.H.S=L.HS.=1.129080309.
– pi-π
Dec 30, 2016 at 15:55

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?

• 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?
– pi-π
Dec 30, 2016 at 16:12
• @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. Dec 30, 2016 at 16:15
• But I have some queries that I want to ask you personally, and get some help.?
– pi-π
Dec 30, 2016 at 16:22

Hint:

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

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

• @ 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 $30$.
– pi-π
Jan 28, 2017 at 9:33
• @NeWtoN, Not sure if I have understood u correctly. Jan 28, 2017 at 9:34
• Hello sir, did you see the question. @lab bhattacharjee
– pi-π
Jan 28, 2017 at 9:41