Question

Given that y=10/sinx+under root3 cosx minimum value of y

Answer 1

it is given that, y = 10/(sinx + √3cosx)

we have to find minimum value of y.

here you see, y $\propto$ 1/(sinx + √3cosx) , right ?

so, for minimum value of y, (sinx + √3cosx) should be maximum.

means, first of all we have to find maximum value of (sinx + √3cosx).

let's resolve the expression.

sinx + √3cosx

= 2(1/2 sinx + √3/2cosx)

= 2(cos60° sinx + sin60° cosx)

from sin(A + B) = sinA.cosB + cosA.sinB

= 2sin(x + 60°)

as we know, -1≤ sine function ≤ 1

so, -1 ≤ sin(x + 60°) ≤ 1

or, -2 ≤ 2sin(x + 60°) ≤ 2

or, -2 ≤ sinx + √3cosx ≤ 2

hence, it is clear that maximum value of (sinx + √3cosx) = 2

so, minimum value of y = 10/2 = 5

Answer 2

Answer:

Explanation:

this pic may help