# Evaluating $\sum_{r=1}^{89} \frac{1}{1+\tan^3 r}$

$$\sum_{r=1}^{89} \frac{1}{1+\tan^3 r }$$ 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!

• You could try to work out something with $$1+\tan^3r=\frac{\cos^3r+\sin^3r}{\cos^3r}=\frac{\cos^2r-\cos r\sin r+\sin^2r}{\cos^3r}=\frac{1-\cos r\sin r}{\cos^3r}$$ Aug 7, 2021 at 8:33
• @DonAntonio The second equality is false. Aug 7, 2021 at 9:00
• @jjagmath Yep: $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ and the factor $\;(\cos r+\sin r)\;$ is missing. May the OP fulfill this. Thanks. Aug 7, 2021 at 9:42

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 $$S=\displaystyle \boxed{{\frac{89}{2}}}$$