JEE Main 2021 — Mathematics Calculus

We have to find $l=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\mathrm{tan}(\pi {\mathrm{cos}}^{2}\theta )}{\mathrm{sin}(2\pi {\mathrm{sin}}^{2}\theta )}$l=θ→0lim​sin(2πsin2θ)tan(πcos2θ)​

Using ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1,$sin2θ+cos2θ=1, we get

$$\Rightarrow l=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\mathrm{tan}(\pi (1-{\mathrm{sin}}^{2}\theta ))}{\mathrm{sin}(2\pi {\mathrm{sin}}^{2}\theta )}$$⇒l=θ→0lim​sin(2πsin2θ)tan(π(1−sin2θ))​

$$\Rightarrow l=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\mathrm{tan}(\pi -\pi {\mathrm{sin}}^{2}\theta )}{\mathrm{sin}(2\pi {\mathrm{sin}}^{2}\theta )}$$⇒l=θ→0lim​sin(2πsin2θ)tan(π−πsin2θ)​

Using $\mathrm{tan}(\pi -\theta )=-\mathrm{tan}\theta ,$tan(π−θ)=−tanθ, we get

$$\Rightarrow l=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{-\mathrm{tan}(\pi {\mathrm{sin}}^{2}\theta )}{\mathrm{sin}(2\pi {\mathrm{sin}}^{2}\theta )}$$⇒l=θ→0lim​sin(2πsin2θ)−tan(πsin2θ)​

$$\Rightarrow l=\underset{\theta \rightarrow 0}{\mathrm{lim}}-(\frac{\mathrm{tan}(\pi {\mathrm{sin}}^{2}\theta )}{\pi {\mathrm{sin}}^{2}\theta })(\frac{2\pi {\mathrm{sin}}^{2}\theta }{\mathrm{sin}(2\pi {\mathrm{sin}}^{2}\theta )})\times \frac{1}{2}$$⇒l=θ→0lim​−(πsin2θtan(πsin2θ)​)(sin(2πsin2θ)2πsin2θ​)×21​

Using, \underset{x\rightarrow 0}{\mathrm{lim}}(\frac{\mathrm{sin}x}{x})=1&\underset{x\rightarrow 0}{\mathrm{lim}}(\frac{\mathrm{tan}x}{x})=1, we get

$$l=-\frac{1}{2}.$$l=−21​.