Question

Prove that $tan^{-1}(\frac{1}{2}) + tan^{-1}(\frac{1}{5}) + tan^{-1}(\frac{1}{8})= \frac{\pi}{4}$.

Answer 1

Solution :

i ) Let tan^-1 ( 1/2 ) = A

=> tanA = 1/2

ii ) tan^-1 ( 1/5 ) = B

=> tanB = 1/5

iii ) tan^-1 ( 1/8 ) = C

=> tanC = 1/8

iv ) tan ( A + B )

= ( tanA + tanB )/( 1 - tanAtanB )

= [ 1/2 + 1/5 ]/[ 1 - ( 1/2 )( 1/5 ) ]

= [ (5+2)/10 ]/[(10-1)/10]

= 7/9

v ) tan ( A + B + C )

= [ tan(A+B) + tanC ]/[1-tan(A+B)tanC]

= [7/9 + 1/8]/[1 - (7/9)(1/8)]

= [(56+9)/72]/[(72-7)/72]

= 65/65

= 1

= tan(π/4)

Therefore,

A + B + C = π/4

tan^-1(1/2) + tan^-1(1/4) + tan^-1(1/8) = π/4

••••

Answer 2

Answer:

Step-by-step explanation:

Formula used:

${tan}^{-1}A+{tan}^{-1}B={tan}^{-1}[\frac{A+B}{1-AB}]$

${tan}^{-1}(\frac{1}{2})+{tan}^{-1}(\frac{1}{5})+{tan}^{-1}\\(\frac{1}{8})$

$={tan}^{-1}[\frac{\frac{1}{2}+\frac{1}{5}}{1- \frac{1}{2}.\frac{1}{5}}]+{tan}^{-1}(\frac{1}{8})$

$={tan}^{-1}[\frac{\frac{7}{10}}{1- \frac{1}{10}}]+{tan}^{-1}(\frac{1}{8})$

$={tan}^{-1}[\frac{\frac{7}{10}}{\frac{9}{10}}]+{tan}^{-1}(\frac{1}{8})$

$={tan}^{-1}(\frac{7}{9})+{tan}^{-1}(\frac{1}{8})$

$={tan}^{-1}[\frac{\frac{7}{9}+ \frac{1}{8}}{1- \frac{7}{9}.\frac{1}{8}}]$

$={tan}^{-1}[\frac{\frac{65}{72}}{1-\frac{7}{72}}]$

$={tan}^{-1}[\frac{\frac{65}{72}}{\frac{65}{72}}]$

$={tan}^{-1}[1]=\pi/4$