Question

If tan (A – B) = 1/

3

and tan (A + B) = √3 , then the value of A and B, respectively are​

Answer 1

Given,

$\longrightarrow \tan(A-B)=\dfrac{1}{3}$

$\longrightarrow A-B=\tan^{-1}\left(\dfrac{1}{3}\right)$

And,

$\longrightarrow \tan(A+B)=\sqrt3$

$\longrightarrow A+B=\tan^{-1}\sqrt3$

Then value of A is,

$\longrightarrow A=\dfrac{\tan^{-1}\sqrt3+\tan^{-1}\left(\dfrac{1}{3}\right)}{2}$

We know that,

• $\tan^{-1}a\pm\tan^{-1}b=\tan^{-1}\left(\dfrac{a\pm b}{1\mp ab}\right)$

Then,

$\longrightarrow A=\dfrac{1}{2}\tan^{-1}\left(\dfrac{\sqrt3+\frac{1}{3}}{1-\sqrt3\times\frac{1}{3}}\right)$

$\longrightarrow\underline{\underline{A=\dfrac{1}{2}\tan^{-1}\left(\dfrac{3\sqrt3+1}{3-\sqrt3}\right)}}$

And, value of B is,

$\longrightarrow B=\dfrac{\tan^{-1}\sqrt3-\tan^{-1}\left(\dfrac{1}{3}\right)}{2}$

$\longrightarrow B=\dfrac{1}{2}\tan^{-1}\left(\dfrac{\sqrt3-\frac{1}{3}}{1+\sqrt3\times\frac{1}{3}}\right)$

$\longrightarrow\underline{\underline{B=\dfrac{1}{2}\tan^{-1}\left(\dfrac{3\sqrt3-1}{3+\sqrt3}\right)}}$