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  • The sum of possible values of x for <img alt="\tan ^{-1}\left ( x+1 \right )+\cot ^{-1}\left ( \frac{1}{x-1} \right )=\tan ^{-1}\left ( \frac{8}{31} \right )" src="/latex-image/?%5Ctan%20%5E%7B-1%7D%5Cleft%20%28%20x&plus;1%20%5Cright%20%29&plus

The sum of possible values of x for \tan ^{-1}\left ( x+1 \right )+\cot ^{-1}\left ( \frac{1}{x-1} \right )=\tan ^{-1}\left ( \frac{8}{31} \right ) is :
 
Option: 1 -\frac{30}{4}
Option: 2 -\frac{33}{4}
Option: 3 -\frac{31}{4}
Option: 4 -\frac{32}{4}

Answers (1)

best_answer

\begin{aligned} &\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}\\ &\text { Taking tangent both sides :- }\\ &\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}\\ &\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}\\ &\Rightarrow 4 x^{2}+31 x-8=0\\ &\Rightarrow x =-8, \frac{1}{4} \end{aligned}

\begin{aligned} & \text { But, if } x=\frac{1}{4} \\ &\tan ^{-1}(x+1) \in\left(0, \frac{\pi}{2}\right) \\ &\& \cot ^{-1}\left(\frac{1}{x-1}\right) \in\left(\frac{\pi}{2}, \pi\right) \end{array}

\Rightarrow LHS >\frac{\pi}{2} \qquad\& \qquad RHS <\frac{\pi}{2}

(Not possible)

Hence, x = -8

Posted by

Suraj Bhandari

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