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

2 Answers
Oct 20, 2016

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)#

Oct 20, 2016

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)#