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?

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?
