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# Solve for x

Question:

Solve for x:

$\tan ^{-1}(\mathrm{x}+1)+\tan ^{-1}(\mathrm{x}-1)=\tan ^{-1} \frac{8}{31}$

Solution:

To find: value of x

Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$

Given: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}$

$\mathrm{LHS}=\tan ^{-1}\left(\frac{\mathrm{x}+1+\mathrm{x}-1}{1-\{(\mathrm{x}+1) \times(\mathrm{x}-1)\}}\right)$

$=\tan ^{-1} \frac{2 x}{1-\left(x^{2}-x+x-1\right)}$

$=\tan ^{-1} \frac{2 x}{2-x^{2}}$

Therefore, $\tan ^{-1} \frac{2 \mathrm{x}}{2-\mathrm{x}^{2}}=\tan ^{-1} \frac{8}{31}$

Taking tangent on both sides, we get

$\frac{2 x}{2-x^{2}}=\frac{8}{31}$

$\Rightarrow 62 x=16-8 x^{2}$

$\Rightarrow 8 x^{2}+62 x-16=0$

$\Rightarrow 4 x^{2}+31 x-8=0$

$\Rightarrow 4 x^{2}+32 x-x-8=0$

$\Rightarrow 4 x \times(x+8)-1 \times(x+8)=0$

$\Rightarrow(4 x-1) \times(x+8)=0$

$\Rightarrow x=\frac{1}{4}$ or $x=-8$

Therefore, $x=\frac{1}{4}$ or $x=-8$ are the required values of $x$.