Y =2/(sinθ + √3cosθ)
for solving this question, you should understand one thing
- ≤ asinx + bcosx ≤
So, -√(1 + √3²) ≤ (sinθ + √3cosθ) ≤ √(1 + √3²)
-2 ≤ (sinθ + √3cosθ) ≤ 2
So, minimum value of (sinθ + √3cosθ) = -2
Maximum value of (sinθ + √3cosθ) = 2
For getting minimum value of y , we have to use maximum value of (sinθ + √3cosθ) .
So, minimum value of y = 2/-2 = 1
Hence, =1