If 2cosec²θ + 3cot²θ = 17, then the value of 'θ' when 0° ≤ θ ≤ 90° is:

Correct — B. Turn the cosec term into a cot term with the Pythagorean identity, then solve for cot θ.

cosec²θ = 1 + cot²θ
2(1 + cot²θ) + 3cot²θ = 17
2 + 5cot²θ = 17, so cot²θ = 3
θ lies in the first quadrant, so cot θ = √3

cot θ = √3 at θ = 30° → option (b).

Substituting back: cosec²30° = 4 and cot²30° = 3, so 2(4) + 3(3) = 8 + 9 = 17.

Why the others are wrong

• (a)
  45° — At 45° cosec²θ = 2 and cot²θ = 1, so the expression comes to 2(2) + 3(1) = 7, not 17.
• (c)
  75° — At 75° cot θ is about 0.27, so cot²θ ≈ 0.07 and cosec²θ ≈ 1.07. The expression is roughly 2.4, nowhere near 17.
• (d)
  90° — At 90° cot θ = 0 and cosec θ = 1, so the expression is 2(1) + 3(0) = 2 — its smallest value anywhere on the stated range.

One identity does all the work: cosec²θ = 1 + cot²θ. Rewriting the cosec term leaves a single unknown, cot²θ, and the equation becomes linear in it.

That is the standard move whenever an equation mixes two co-functions of the same angle. Convert to whichever one appears more often, solve for its square, then read the angle off the standard table.

The stated range 0° ≤ θ ≤ 90° is what forces a single answer. Without it, cot²θ = 3 would also admit 150°, where cot θ = −√3.

Key facts

• cosec²θ − cot²θ = 1.
• cot 30° = √3, cot 45° = 1, cot 60° = 1 ⁄ √3 and cot 90° = 0.
• 2cosec²θ + 3cot²θ simplifies to 2 + 5cot²θ.

Common traps

• Writing cosec²θ = 1 − cot²θ, which is the sine identity misremembered.
• Solving cot²θ = 3 and answering 60°, the angle where tan θ = √3.
• Stopping at cot θ = √3 without converting that to a degree measure.

No directly related past PYQ was found.