I know
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ but how can I prove
$$4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{70}+\tan^{-1}\frac{1}{99}=\frac{\pi}{4}$$
I tried it, but I am getting large numbers using the above formula. I think there must be some trick to follow. Regards
Set $\tan \alpha = \frac 15$. Use the duplication formula $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$ twice to obtain $\tan 4 \alpha = \frac{120}{119}$. Hence, $\tan{4 \alpha} - 1 = \frac{1}{119}$ and $\tan\left(4 \alpha - \frac \pi 4\right) = \frac{\frac 1{119}}{ (1 + \frac{120}{119})} = \frac 1{239}$ from the fact that $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. Hence, $\arctan \frac 1{239} = 4 \alpha - \frac \pi 4$.
On the other hand, $\arctan \frac 1{70} - \arctan \frac 1{99}$ can be calculated from the formula you gave at first, and it gives $\arctan \frac 1{239}$.
Hence, $\arctan \frac 1{70} - \arctan \frac 1{99} = 4 \arctan \frac 15 - \frac \pi 4$. Rearrange to get the desired result.
Note : the fact that $4 \alpha - \frac \pi 4 = \arctan \frac 1{239}$ is a very special one, called Machin's formula, named after John Machin, who used it to calculate digits of $\pi$ way back in $1706$. More about that on : https://en.wikipedia.org/wiki/Machin-like_formula. There are some really amazing formulas on that page.