JEE Main 2021 — Mathematics Trigonometry

Given ${\mathrm{tan}}^{-1}(x+1)+{\mathrm{cot}}^{-1}(\frac{1}{x-1})={\mathrm{tan}}^{-1}\frac{8}{31}$tan−1(x+1)+cot−1(x−11​)=tan−1318​

$$\Rightarrow {\mathrm{tan}}^{-1}(x+1)+{\mathrm{tan}}^{-1}(x-1)={\mathrm{tan}}^{-1}\frac{8}{31},(\therefore {\mathrm{tan}}^{-1}x={\mathrm{cot}}^{-1}\frac{1}{x},x>0)$$⇒tan−1(x+1)+tan−1(x−1)=tan−1318​,(∴tan−1x=cot−1x1​,x>0)

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{x+1+x-1}{1-(x+1)(x-1)})={\mathrm{tan}}^{-1}\frac{8}{31}$$⇒tan−1(1−(x+1)(x−1)x+1+x−1​)=tan−1318​

$$\Rightarrow {\mathrm{tan}}^{-1}[\frac{2x}{1-(x+1)(x-1)}]={\mathrm{tan}}^{-1}\frac{8}{31}$$⇒tan−1[1−(x+1)(x−1)2x​]=tan−1318​

Taking tangent both the sides we get,

$$\frac{(x+1)+(x-1)}{1-({x}^{2}-1)}=\frac{8}{31}$$1−(x2−1)(x+1)+(x−1)​=318​

$$\Rightarrow \frac{2x}{2-{x}^{2}}=\frac{8}{31}$$⇒2−x22x​=318​

$$\Rightarrow 4{x}^{2}+31x-8=0$$⇒4x2+31x−8=0

$$\Rightarrow x=-8,\frac{1}{4}$$⇒x=−8,41​

But, if $x=\frac{1}{4}$x=41​

$${\mathrm{tan}}^{-1}(x+1)\in (0,\frac{\pi }{2})$$tan−1(x+1)∈(0,2π​)

and ${\mathrm{cot}}^{-1}(\frac{1}{x-1})\in (\frac{\pi }{2},\pi )$cot−1(x−11​)∈(2π​,π)

$\Rightarrow \mathrm{LHS}>\frac{\pi }{2}$⇒LHS>2π​ and $\mathrm{RHS}<\frac{\pi }{2}$RHS<2π​ (Not possible)

Hence, $x=-8=-\frac{32}{4}.$x=−8=−432​.