JEE Main 2022 — Mathematics Calculus

Given, $\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}{\mathrm{tan}}^{2}x[\sqrt{2{\mathrm{sin}}^{2}x+3\mathrm{sin}x+4}-\sqrt{{\mathrm{sin}}^{2}x+6\mathrm{sin}x+2}]$x→2π​lim​tan2x[2sin2x+3sinx+4​−sin2x+6sinx+2​]

$$=\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}{\mathrm{tan}}^{2}x\frac{[{(\sqrt{2{\mathrm{sin}}^{2}x+3\mathrm{sin}x+4})}^{2}-{(\sqrt{{\mathrm{sin}}^{2}x+6\mathrm{sin}x+2})}^{2}]}{\sqrt{2{\mathrm{sin}}^{2}x+3\mathrm{sin}x+4}+\sqrt{{\mathrm{sin}}^{2}x+6\mathrm{sin}x+2}}$$=x→2π​lim​tan2x2sin2x+3sinx+4​+sin2x+6sinx+2​[(2sin2x+3sinx+4​)2−(sin2x+6sinx+2​)2]​

$$=\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}\frac{{\mathrm{tan}}^{2}x[{\mathrm{sin}}^{2}x-3\mathrm{sin}x+2]}{\sqrt{9}+\sqrt{9}}$$=x→2π​lim​9​+9​tan2x[sin2x−3sinx+2]​

$$=\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}\frac{{\mathrm{tan}}^{2}x(\mathrm{sin}x-1)(\mathrm{sin}x-2)}{6}$$=x→2π​lim​6tan2x(sinx−1)(sinx−2)​

$$=\frac{1}{6}\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}{\mathrm{tan}}^{2}x(1-\mathrm{sin}x)$$=61​x→2π​lim​tan2x(1−sinx)

$$=\frac{1}{6}\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}x(1-\mathrm{sin}x)}{(1-\mathrm{sin}x)(1+\mathrm{sin}x)}=\frac{1}{12}$$=61​x→2π​lim​(1−sinx)(1+sinx)sin2x(1−sinx)​=121​