Trigonometry basic proof problem

Question

I tried to solve but I think I am making a mistake somewhere. Could you help me solving this?

Show that $\arctan(\frac{1}{4}) + \arctan(\frac{3}{5})= \frac{\pi}{4}$

Hence or otherwise, find the value of $\arctan(4)+ \arctan(\frac{3}{5})$

So i did $\tan(x)= \dfrac{1}{4}$ and $\tan(y)= \dfrac{3}{5}$ then, I added them which is $\dfrac{17}{20}= 0.85$ and this gives $40$ degrees.

I am confused.

Answer 1

Since $\tan x=\frac14$ and $\tan y=\frac35$,$$\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=\frac{\frac{17}{20}}{1-\frac3{20}}=1.$$Can you take it from here?

Answer 2

HINT

Recall that

$$\arctan x +\arctan y=\arctan\frac{x+y}{1-xy}$$

take a look here for the proof.