Prove \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{31}{17}\)
To prove \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{31}{17}\)
LHS= \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}\)
= \(\tan^{-1} \frac{\frac{2.1}{2}}{1-(\frac{1}{2})^2}+\tan^{-1}\frac{1}{7} \ [2tan^{-1x}=\tan^{-1}\frac{2x}{1-x^2}]\)
\(\tan^{-1}\frac{1}{(\frac{3}{4})}+\tan^{-1}\frac{1}{7}\)
\(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{7}\)
= \(\tan^{-1}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}\)
=\(\tan^{-1}\frac{\frac{(28+3)}{21}}{\frac{(21-4)}{21}}\)
= \(\tan^{-1}\frac{31}{17}\)=RHS
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tanβ1x + Tanβ1y = Ο + tanβ1 (x+y/ 1-xy), if xy > 1
Tanβ1x + Tanβ1y = tanβ1 (x+y/ 1-xy), if xy < 1
Tanβ1x + Tanβ1y = tanβ1 (x+y/ 1-xy), if xy < 1
Tanβ1x + Tanβ1y = -Ο + tanβ1 (x+y/ 1-xy), if xy > 1
= x, if xβ[βΟ/2, Ο/2]
= Οβx, if xβ[Ο/2, 3Ο/2]
=β2Ο+x, if xβ[3Ο/2, 5Ο/2] And so on.
= βx, β[βΟ,0]
= x, β[0,Ο]
= 2Οβx, β[Ο,2Ο]
=β2Ο+x, β[2Ο,3Ο]
= x, (βΟ/2, Ο/2)
= xβΟ, (Ο/2, 3Ο/2)
= xβ2Ο, (3Ο/2, 5Ο/2)
