Answer :
Answer:
a=4 and b = 0
Step-by-step explanation:
Given : [tex]\dfrac{\sqrt 3 -1}{\sqrt 3+1}+\dfrac{\sqrt 3+1}{\sqrt 3-1}=a+\sqrt 3 b[/tex]
To find, The value of a and b
Solution,
Solving LHS of the given equation,
[tex]\dfrac{(\sqrt 3-1)^2+(\sqrt3+1)^2}{(\sqrt 3+1)(\sqrt 3-1)}=a+\sqrt 3 b[/tex]
Since,
[tex](a-b)^2=a^2+b^2-2ab\\\\(a+b)^2=a^2+b^2+2ab\\\\(a-b)(a+b)=a^2+b^2[/tex]
So,
[tex]\dfrac{3+1-2\sqrt 3+3+1+2\sqrt 3}{3-1}=a+\sqrt 3 b\\\\4=a+\sqrt 3 b[/tex]
or
[tex]4+0=a+\sqrt 3 b[/tex]
On comparing we get :
a = 4 and b = 0