I have some trouble with my homework. The problem is to simplify the following trigonometric expression: $$ \frac{\sin150^{\circ}-\cos240^{\circ}}{\cot730^{\circ}\cot800^{\circ}+\tan730^{\circ}\tan800^{\circ}} $$
I know that $\sin{150^{\circ}} = \sin{\frac{5\pi}{3}} = \frac{1}{2}$ and $\cos240^{\circ} = \cos{\frac{4\pi}{3}}=-\frac{1}{2}$, so $\sin150^{\circ}-\cos240^{\circ} = 1$, and numerator equals $1$. How do I deal with denominator?
Due to periodic properties of sine and cosine we have $\cot730^{\circ} = \frac{1}{\tan730^{\circ}} = \cot10^{\circ} = \frac{1}{\tan10^{\circ}}$ and $\cot800^{\circ} = \frac{1}{\tan800^{\circ}} = \cot80^{\circ} = \frac{1}{\tan80^{\circ}}$. Therefore: $$ \frac{1}{\cot730^{\circ}\cot800^{\circ}+\tan730^{\circ}\tan800^{\circ}} = \frac{1}{\cot10^{\circ}\cot80^{\circ}+\tan10^{\circ}\tan80^{\circ}} $$
Now note that $$\cos10^{\circ} = \cos(-10^{\circ}) = \sin(-10^{\circ} + 90^{\circ}) = \sin80^{\circ},$$ $$\cos80^{\circ} = -\sin(80^{\circ}-90^{\circ})= -\sin(-10^{\circ}) = \sin10^{\circ},$$ so $$ \tan10^{\circ} = \frac{\sin10^{\circ}}{\cos10^{\circ}}= \frac{\cos80^{\circ}}{\sin80^{\circ}} = \cot80^{\circ}, $$ $$ \tan80^{\circ} = \frac{\sin80^{\circ}}{\cos80^{\circ}}= \frac{\cos10^{\circ}}{\sin10^{\circ}} = \cot10^{\circ}, $$ so you have $$ \frac{1}{\cot10^{\circ}\cot80^{\circ}+\tan10^{\circ}\tan80^{\circ}} = \frac{1}{\cot10^{\circ}\tan10^{\circ}+\tan10^{\circ}\cot10^{\circ}} = \frac{1}{2} $$
Let me give you some hints:
$\cot A = \tan (90^{\circ} - A)$ and vice-versa
and $\tan A=\frac{1}{\cot A}$ and vice-versa
$\cot730^{\circ}=\cot10^{\circ} $
$\cot800^{\circ}=\cot80^{\circ}$
Try to figure out the rest by yourself.
