Summation of trigonometry functions

$\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{\sin^2(\theta)=\sin^2(180^{\circ}-\theta)\\ \sin^2(\theta)=\cos^2(90^{\circ}-\theta)\\ \cos^2(\theta)=\cos^2(180^{\circ}-\theta)}}}}$

$\large{\boxed{\sin^2{0^{\circ}}+\sin^2{10^{\circ}}+\sin^2{20^{\circ}}+\sin^2{30^{\circ}}+\ldots+\sin^2{180^{\circ}}}}$ $\large2[\sin^2{0^{\circ}}+\sin^2{10^{\circ}}+\sin^2{20^{\circ}}+\sin^2{30^{\circ}}+\ldots+\sin^2{80^{\circ}}]+ \sin^2{90^{\circ}}$ $\large2[4+\sin^2{0^{\circ}}]+\sin^2{90^{\circ}}$ $\large2[4]+\sin^2{90^{\circ}}=9$ $\rightarrow\sin^2{0^{\circ}}+\sin^2{10^{\circ}}+\sin^2{20^{\circ}}+\sin^2{30^{\circ}}+\ldots+\sin^2{180^{\circ}} ={\boxed{9}}$

$\large{\boxed{\cos^2{0^{\circ}}+\cos^2{10^{\circ}}+\cos^2{20^{\circ}}+\cos^2{30^{\circ}}+\ldots+\cos^2{180^{\circ}}}}$ $\large2[\cos^2{0^{\circ}}+\cos^2{10^{\circ}}+\cos^2{20^{\circ}}+\cos^2{30^{\circ}}+\ldots+\cos^2{80^{\circ}}]+ \cos^2{90^{\circ}}$ $\large2[4+\cos^2{0^{\circ}}]+\cos^2{90^{\circ}}$ $\large2[5]+\cos^2{90^{\circ}}=10$ $\rightarrow\cos^2{0^{\circ}}+\cos^2{10^{\circ}}+\cos^2{20^{\circ}}+\cos^2{30^{\circ}}+\ldots+\cos^2{180^{\circ}} ={\boxed{10}}$ $\large\dfrac{\sin^2{0^{\circ}}+\sin^2{10^{\circ}}+\sin^2{20^{\circ}}+\sin^2{30^{\circ}}+\ldots+\sin^2{180^{\circ}}}{\cos^2{0^{\circ}}+\cos^2{10^{\circ}}+\cos^2{20^{\circ}}+\cos^2{30^{\circ}}+\ldots+\cos^2{180^{\circ}}}=\frac{9}{10}=\frac{A}{B}$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{A^2+B^2=81+100=181}}}}$

Using the basic identity sin^2@+cos^2@=1 ,and transformations... We'll get the ratio as 9/10. Note that the denominator forms Gud enough pairs to add up till 9 + 1 coz of cos^2(180) in the end of the sequence.similarly in the numerator but sin^2(180)=0..hence 9/10.. 9^2+10^2=181.

Simply use sinx = [e^{ix}-e^{-ix}]/2 and the problem becomes a piece of cake.