The value of the limit
lim _(θ → 0) (tan (π cos ²θ ))/(sin (2π sin ²θ )) is equal to :

Solution

Maths sir · 4 steps

Write π cos² θ as π - π sin² θ and the top and bottom share one small angle.

1. Spot the form. At θ = 0 the numerator is tan π = 0 and the denominator is sin 0 = 0. This is a 0/0 form, so we need to rewrite it.
2. Shift the angle. Since cos² θ = 1 - sin² θ, we get tan(π cos² θ) = tan(π - π sin² θ) = -tan(π sin² θ), because tan(π - x) = -tan x.
3. Substitute. Let u = π sin² θ. As θ → 0, u → 0. Then use tan u ≈ u and sin 2u ≈ 2u for small u.

   lim(u → 0) (-tan u)/(sin 2u) = lim(u → 0) (-(tan u)/u)/(2 × (sin 2u)/2u) = -1/2

4. So. The limit is -1/2, which is option (b).

The idea to keep. When a trig argument tends to π instead of 0, shift it with tan(π - x) = -tan x or a similar identity. Then the standard limits (tan x)/x → 1 and (sin x)/x → 1 apply.

Answer from the NTA final answer key. Solution by PrepWiser.