$$\sum_{r=1}^{89} \frac{1}{1+\tan^3 r }$$
where $r$ is in degrees
I tried this a lot using the $a^3+b^3$ identity but I don't seem to be getting anything fruitful 🙁
Can someone please give me a hint? I don't think the $V_n$ method is working here. ($T_r+T_{r-1}$)
Is it related to the $\tan(A+B)$ identity? However I'm not getting anything through that too.
I would be grateful if someone helped. Thanks!
Best Answer
Let $S=\displaystyle\sum_{r=1}^{89} \cfrac{1}{1+\tan^3 r }$
$S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3r}{\cos^3r+\sin^3 r }$
Now Since $\displaystyle\sum_{r=a}^{b}f(r)=\displaystyle\sum_{r=a}^{b}f(a+b-r)$
Using which we get $S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3(90-r)}{\cos^3(90-r)+\sin^3(90-r) }=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r}{\cos^3r+\sin^3 r }$
Therefore $S=\displaystyle\sum_{r=1}^{89} \frac{\cos^3r}{\cos^3r+\sin^3 r} =\displaystyle\sum_{r=1}^{89} \frac{\sin^3r}{\cos^3r+\sin^3 r }$
Therefore $2S=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r+\cos^3r}{\cos^3r+\sin^3 r }=\displaystyle\sum_{r=1}^{89} 1=89$
Therefore $S=\displaystyle \boxed{{\frac{89}{2}}}$