Solve this
Question:

It $\tan ^{-1} \frac{x+1}{x-1}+\tan ^{-1} \frac{x-1}{x}=\tan ^{-1}(-7)$, then the value of $x$ is

(a) 0

(b) $-2$

(c) 1

(d) 2

Solution:

(d) 2

We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$.

$\therefore \tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{x+1}{x-1}+\frac{x-1}{z}}{1-\frac{x+1}{x-1} \times \frac{x-1}{x}}\right)=\tan ^{-1}(-7)$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{x^{2}+x+x^{2}-2 x+1}{x x-1}}{\frac{x^{2}-x-x^{2}+1}{x(x-1)}}\right)=\tan ^{-1}(-7)$

$\Rightarrow \tan ^{-1}\left(\frac{2 x^{2}-x+1}{-x+1}\right)=\tan ^{-1}(-7)$

So, we get

$\frac{2 x^{2}-x+1}{-x+1}=-7$

$\Rightarrow 2 x^{2}-x+1=7 x-7$

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

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

$\Rightarrow(x-2)^{2}=0$

$\Rightarrow x=2$

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