Question

Y=2/(sin theta + √3 cos theta), then the minimum value of y is?

Answer 1

y =2/(sinθ + √3cosθ)
for solving this question, you should understand one thing
-$\bf{\sqrt{a^2+b^2}}$ ≤ asinx + bcosx ≤$\bf{\sqrt{a^2+b^2}}$

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, $\bf{y_{min}}$=1

Answer 2

Answer:

see the attachment.......

Step-by-step explanation: