JEE Main 2022 — Mathematics Trigonometry

Given $x\times y={x}^{2}+{y}^{3}$x×y=x2+y3

Now $(x\times 1)\times 1=x\times (1\times 1)$(x×1)×1=x×(1×1)

$$({x}^{2}+{1}^{3})\times 1=x\times ({1}^{2}+{1}^{3})$$(x2+13)×1=x×(12+13)

$${({x}^{2}+1)}^{2}+{1}^{3}={x}^{2}+{2}^{3}$$(x2+1)2+13=x2+23

$${x}^{4}+1+2{x}^{2}+1={x}^{2}+8$$x4+1+2x2+1=x2+8

$${x}^{4}+{x}^{2}-6=0$$x4+x2−6=0

Let ${x}^{2}=t$x2=t

$${t}^{2}+t-6=0$$t2+t−6=0

$$(t+3)(t-2)=0$$(t+3)(t−2)=0

$t\neq -3$t=−3 so $t=2$t=2 or ${x}^{2}=2$x2=2

Now $2{\mathrm{sin}}^{-1}(\frac{{x}^{4}+{x}^{2}-2}{{x}^{4}+{x}^{2}+2})=2{\mathrm{sin}}^{-1}(\frac{4+2-2}{4+2+2})$2sin−1(x4+x2+2x4+x2−2​)=2sin−1(4+2+24+2−2​)

$$=2{\mathrm{sin}}^{-1}(\frac{1}{2})=\frac{\pi }{6}\times 2=\frac{\pi }{3}$$=2sin−1(21​)=6π​×2=3π​