Question

If $ta{n}^{-1}\frac{x-3}{x-4}+ta{n}^{-1}\frac{x+3}{x+4}=\frac{3}{4}$, then find the value of $x$.

Answer 1

Given $ta{n}^{-1}\left(\frac{x-3}{x-4}\right)+ta{n}^{-1}\left(\frac{x+3}{x+4}\right)=\frac{3}{4}$

Let ${\tan }^{-1}\left(\frac{x-3}{x-4}\right)=a$ and ${\tan }^{-1}\left(\frac{x+3}{x+4}\right)=b$

So we have $\frac{x-3}{x-4}=tana$ and $\frac{x+3}{x+4}=tanb$

The given equation will become $a+b=\frac{3}{4}$ , Now apply $tan$ on both sides

We get $tan(a+b)=\frac{tana+tanb}{1-tana\times tanb}=tan\left(\frac{3}{4}\right)$

By substituting $tana$ and $tanb$ values , we get $\frac{\frac{x-3}{x-4}+\frac{x+3}{x+4}}{1-\frac{x-3}{x-4}\times \frac{x+3}{x+4}}=\frac{(x-3)(x+4)+(x+3)(x-4)}{(x-4)(x+4)+(x-3)(x+3)}=\frac{2x^2-24}{2x^2-25}=\tan \left(\frac{3}{4}\right)$

By solving , we get $x^2=\frac{24-25tan\left(\frac{3}{4}\right)}{2-2tan\left(\frac{3}{4}\right)}$