Question

How do you prove `tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)`?

Answer 1

See explanation...

Explanation:

For any `alpha` and `beta` we have:

`tan(alpha+beta) = (tan(alpha)+tan(beta)) / (1-tan(alpha)tan(beta))`

Let `alpha = tan^(-1)(1/3)` and `beta = tan^(-1)(1/5)`

Note that `alpha in [0, 1)` and `beta in [0, 1)`, so `alpha < tan^(-1) = pi/4` and `beta < pi/4`. Hence `alpha+beta < pi/2` is in the range of `tan^(-1)`

Then we find:

`tan(alpha+beta) = (tan(alpha) + tan(beta))/(1-tan(alpha)tan(beta))`

`color(white)(tan(alpha+beta)) = (1/3 + 1/5)/(1-1/3*1/5)`

`color(white)(tan(alpha+beta)) = (8/15)/(14/15)`

`color(white)(tan(alpha+beta)) = 8/14`

`color(white)(tan(alpha+beta)) = 4/7`

So:

`tan^(-1)(1/3) + tan^(-1)(1/5) = alpha + beta = tan^(-1)(4/7)`

Answer 2

See the proof below
`tan^-1(4/7)=tan^-1(1/3)+tan^-1(1/5)`

Explanation:

Let us have two angles a and b
Then `tan(a+b)=sin(a+b)/cos(a+b)=( sinacosb+sinbcosa)/(cosacosb-sinasinb)`
Divide throughout by `cosacosb`
`tan(a+b)=(((sinacosb)/(cosacosb))+((sinbcosa)/(cosacosb)))/((cosacosb)/(cosacosb)-(sinasinb)/(cosacosb)`
Simplifyng we get
`tan(a+b)=(tana +tanb)/(1-tanatanb)`
Here we have `tana=1/3` and `tanb=1/5`
so `tan(a+b)=(1/3+1/5)/(1-(1/3*1/5))=(8/15)/(14/15)=8/14=4/7`
`tan^-1(4/7)=tan^-1(1/3)+tan^-1(1/5)`