Question

Prove that $tan^{-1}(\frac{1}{7}) + tan^{-1}(\frac{1}{13}) - tan^{-1}(\frac{2}{9})= 0$.

Answer 1

Solution :

i ) Let tan^-1( 1/7 ) = A

=> tanA = 1/7

ii ) tan^-1( 1/13 ) = B

=> tanB = 1/13

iii ) tan^-1( 2/9 ) = C

=> tanC = 2/9

iv ) tan( A + B )

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

= [ 1/7 + 1/13 ]/[ 1 - (1/7)(1/13) ]

= [(13+7)/91]/[(91-1)/91]

= 20/90

= 2/9

v ) tan ( A + B - C )

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

= [ 2/9 - 2/9 ]/[ 1 + (2/9)(2/9)]

= ( 0 ) /[ ( 81-4 )/81 ]

= ( 0 )/(77/81)

= 0

= tan 0°

Therefore ,

A + B - C = 0

tan^-1(1/7)+ tan^-1(1/13) - tan^-1(2/9) = 0

••••

Answer 2

Answer:

Step-by-step explanation:

Formula used:

${tan}^{-1}A+{tan}^{-1}B={tan}^{-1}(\frac{A+B}{1-AB})$

${tan}^{-1}\frac{1}{7}+{tan}^{-1}\frac{1}{13}-{tan}^{-1}\frac{2}{9}$

$={tan}^{-1}(\frac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{7}.\frac{1}{13}})-{tan}^{-1}\frac{2}{9}$

$={tan}^{-1}(\frac{\frac{20}{91}}{1-\frac{1}{91}})-{tan}^{-1}\frac{2}{9}$

$={tan}^{-1}(\frac{\frac{20}{91}}{\frac{90}{91}})-{tan}^{-1}\frac{2}{9}$

$={tan}^{-1}\frac{20}{90}-{tan}^{-1}\frac{2}{9}$

$={tan}^{-1}\frac{2}{9}-{tan}^{-1}\frac{2}{9}\\=0$