lim _(x → 1/√2) (sin (cos ⁻¹x) - x)/(1 - tan (cos ⁻¹x)) is equal to :

Solution

Maths sir · 4 steps

Put θ = cos⁻¹ x and the limit turns into a plain trig expression that cancels.

1. Substitute. Let θ = cos⁻¹ x, so x = cos θ. As x → 1/√2, θ → π/4, and the expression becomes (sin θ - cos θ)/(1 - tan θ). At θ = π/4 both top and bottom are 0, which is the 0/0 form.
2. Rewrite the denominator. 1 - tan θ = (cos θ - sin θ)/(cos θ). The bracket cos θ - sin θ is the negative of the numerator sin θ - cos θ.

   (sin θ - cos θ)/(1 - tan θ) = ((sin θ - cos θ)cos θ)/(cos θ - sin θ) = -cos θ

3. Check with L'Hopital. Differentiating top and bottom in θ gives (cos θ + sin θ)/(-sec² θ). At θ = π/4 this is √2/-2 = -1/√2, the same value.
4. So. The limit is -cos π/4 = -1/√2, which is option (d).

The idea to keep. When an inverse trig function sits inside a trig function, substitute θ for the inverse first. The limit usually becomes a simple trig ratio you can factor and cancel.

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