Trigonometry & Inverse Trigonometry Question - None

Question

Statement I: $\csc^{-1}\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) > \sec^{-1}\left(\frac{1}{2} + \frac{1}{2}\right)$
Statement II: $\csc^{-1}x > \sec^{-1}x$ if $1 < x < \sqrt{2}$

Accepted Answer

Step 1: Evaluate Statement I

Left side: csc⁻¹(1/√2 - 1/√2) = csc⁻¹(0)

Since csc⁻¹ has range [-π/2, π/2]\{0} and csc(x) never equals 0 for any x in this range, csc⁻¹(0) is undefined.

Right side: sec⁻¹(1/2 + 1/2) = sec⁻¹(1) = 0 (since sec(0) = 1)

Since the left side is undefined, the inequality cannot be evaluated. Statement I is FALSE.

Step 2: Evaluate Statement II

For 1 < x ≤ √2, we need to verify: csc⁻¹(x) > sec⁻¹(x)

Let α = csc⁻¹(x), so csc(α) = x, where α ∈ (0, π/2]

Then sin(α) = 1/x

Let β = sec⁻¹(x), so sec(β) = x, where β ∈ [0, π/2)

Then cos(β) = 1/x

For 1 < x ≤ √2: We have 1/√2 ≤ 1/x < 1

Since sin(α) = 1/x and cos(β) = 1/x with 1/√2 ≤ 1/x < 1:

For 1/√2 ≤ y < 1: arcsin(y) > arccos(y) because sin⁻¹(y) + cos⁻¹(y) = π/2, and when 1/√2 ≤ y < 1, arcsin(y) > π/4 while arccos(y) < π/4

Since sin(α) = cos(β) = 1/x and arcsin is increasing:

α > β, therefore csc⁻¹(x) > sec⁻¹(x). Statement II is TRUE.

Step 3: Conclusion

Statement I is False; Statement II is True. The answer is option (d).

∴ Answer: d

Question:

Step-by-Step Solution

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