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

Question

Brooklyn Anderson · Accepted Answer

See 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)#

Dylan Arnold · Answer

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

Victoria Adebayo · Answer

undefined To prove $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{4}{7}\right)$, we can use the identity:

$$\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$$

Substituting $a = \frac{1}{3}$ and $b = \frac{1}{5}$ into this identity, we get:

$$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \cdot \frac{1}{5}}\right)$$

Simplify the expression inside the $\tan^{-1}$ function:

$$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{5 + 3}{15}}{1 - \frac{1}{15}}\right)$$

$$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{8}{15}}{\frac{14}{15}}\right)$$

$$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{8}{14}\right)$$

$$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{4}{7}\right)$$

Thus, $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{4}{7}\right)$.