Question

Y is equal to 2 upon sin theta + root 3 cos theta then the minimum value of y is

Answer 1

Answer:

here is ur answere

Step-by-step explanation:

y =2/(sinθ + √3cosθ)

for solving this question, you should understand one thing

-\bf{\sqrt{a^2+b^2}}

a

2

+b

2

≤ asinx + bcosx ≤\bf{\sqrt{a^2+b^2}}

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}}y

min

=1

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Answer 2

Answer:

√3 - 1

Step-by-step explanation:

y = 2 / sinx + √3 cos x

sinx + √3 cos x ≠ 0

if y should be min

                                  sinx + √3 cos x should be max

so max value of sinx is 1 and cosx is also 1

so 1 + √3

so min value is  y is 2 / 1 + √3

                       = √3 - 1 ( rationalize )