The value of the limit $\operatorname{Lim}_{\theta \rightarrow 0} \frac{\tan \left(\pi \cos ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$ is equal to:

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Limits

The value of the limit

Lim_{θ \to 0} \frac{t a n ( π c o s ^{2} θ )}{s i n ( 2 π s i n ^{2} θ )}

is equal to:

Answer: (B)

Using

cos^{2} θ = 1 - sin^{2} θ

, the numerator is

tan (π - π sin^{2} θ) = - tan (π sin^{2} θ)

. Writing

u = sin^{2} θ

, so that

u \to 0

θ \to 0

, the expression becomes

- \frac{tan ( π u )}{sin ( 2 π u )} = - \frac{1}{2} \cdot \frac{tan ( π u )}{π u} \cdot \frac{2 π u}{sin ( 2 π u )}

Both ratios tend to

1

, so the limit is

- \frac{1}{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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