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.

The mistake was that you added the values of $\tan x$ and $\tan y$, but to find the tangent of a sum, you would use the formulas posted below. — bjcolby15, Apr 21 '18 at 14:08

score 1 · Accepted Answer · answered Apr 21 '18 at 13:42

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?

answered Apr 21 '18 at 13:42

José Carlos Santos

427,504

then how do i find find the value of arctan(4)+ arctan(5/3)? – QuantumCasanova Apr 21 '18 at 13:52
@QuantumCasanova If $\tan\alpha=\frac14$, then $\tan\left(\frac\pi2-\alpha\right)=\cot\alpha=\frac1{\tan\alpha}=4$. Again, can you take it from here? – José Carlos Santos Apr 21 '18 at 13:54
I would be very pleased if you continued :) – QuantumCasanova Apr 21 '18 at 14:04
If $\tan\alpha=4$, then $\tan\left(\frac\pi2-\alpha\right)=\frac14$ and if $\tan\beta=\frac53$, then $\tan\left(\frac\pi2-\beta\right)=\frac35$. Therefore, $\left(\frac\pi2-\alpha\right)+\left(\frac\pi2-\beta\right)=\frac\pi4$, which means that $\alpha+\beta=\frac{3\pi}4$. – José Carlos Santos Apr 21 '18 at 14:08

score 0 · Answer 2 · answered Apr 21 '18 at 13:37

0

HINT

Recall that

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

take a look here for the proof.

answered Apr 21 '18 at 13:37

user

154,566

Trigonometry basic proof problem

2 Answers2