If 2cosec²θ + 3cot²θ = 17, then the value of 'θ' when 0° ≤ θ ≤ 90° is:
30°
Given: \(2\csc^2\theta + 3\cot^2\theta = 17\).
Approach: Test the standard angles from the options.
At \(\theta = 30^\circ\):
\(\csc 30^\circ = 2 \Rightarrow \csc^2 30^\circ = 4\)
\(\cot 30^\circ = \sqrt{3} \Rightarrow \cot^2 30^\circ = 3\)
Substituting: \(2(4) + 3(3) = 8 + 9 = 17\) ✓
The equation is satisfied.
Verification (other options fail): At \(\theta = 45^\circ\), \(2(2) + 3(1) = 7 \neq 17\). At \(\theta = 60^\circ\), \(2(4/3) + 3(1/3) = 11/3 \neq 17\).
Hence, the correct answer is \(\theta = 30^\circ\).
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