JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The sum of possible values of \(x\) for \(\tan ^{-1}( x +1)+\cot ^{-1}\left(\frac{1}{ x -1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)\) is
- A \(-\frac{32}{4}\)
- B \(-\frac{31}{4}\)
- C \(-\frac{30}{4}\)
- D \(-\frac{33}{4}\)
Answer & Solution
Correct Answer
(A) \(-\frac{32}{4}\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}\) Taking tangent both sides :- \(\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}\) \(\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}\) \(\Rightarrow 4 x^{2}+31 x-8=0\)…
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