Find $S=(\tan 1^o)+(\tan 2^o)+(\tan 3^o)+(\tan 4^o) +(\tan 5^o)+(\tan 6^o)+(\tan 7^o)+(\tan 8^o)+(\tan 9^o)+(\tan 10^o)$.
My attempt:
$\cos x+\cos 2x+\cos 3x+ . . . +\cos nx =\frac{\sin(n+1/2)x-\sin x/2 }{2\sin x/2}$
$\sin x+\sin 2x+\sin 3x+ . . . +\sin nx =\frac{\cos x/2 - \cos (n+ 1/2)x}{2 \sin x/2} $
Since angles are small we use term by term division which gives:
$$\tan x + \tan 2x+\tan 3x + . . . + \tan nx=\frac{\cos x/2 - \cos (n+ 1/2)x}{\sin(n+1/2)x-\sin x/2 }$$
Putting $x=1$ and $n=10$ gives $s=0.9629..$
Wolfram alpha gives $s=0.9653..$
Is my calculations correct?
What you are writing is not correct since, in essence, you write that $$\frac a b+\frac cd +\frac ef=\frac{a+c+e}{b+d+f}$$ which is not true. For example $$\frac 12+\frac 34 +\frac 56=\frac{25}{12} \qquad \text{while} \qquad \frac{1+3+5}{2+4+6}=\frac{3}{4}$$
However, since you notice that all angles are small, you could approximate the summation using that, for small $x$ $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\tan(ix)=i x+\frac{i^3 x^3}{3}+\frac{2 i^5 x^5}{15}+O\left(x^7\right)$$ $$\sum_{i=1}^n \tan(ix)=x\sum_{i=1}^n i+\frac{ x^3}{3}\sum_{i=1}^n i^3+\frac{2 x^5}{15}\sum_{i=1}^n i^5+\cdots$$ and use Faulhaber's formulae, namely $$\sum_{i=1}^n i=\frac{1}{2} n (n+1)$$ $$\sum_{i=1}^n i^3=\frac{1}{4} n^2 (n+1)^2$$ $$\sum_{i=1}^n i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right)$$ Using $n=10$ and $x=\frac \pi {180}$, the approximation of the sum would be $$\frac{11 \pi }{36}+\frac{121 \pi ^3}{699840}+\frac{8833 \pi ^5}{56687040000}\approx 0.9653396545$$ while the exact sum would be $\approx 0.9653401411$.