Question

Ksin^2(x)+[cosec^2(x)/k]=2,x belongs to (0,π/2),then find the value of cos^2(x)+5sinx.cosx+6sin^2(x) is

Answer 1

Note that $k\sin^2x+\frac{\csc^2(x)}{k}$ attains a minimum value of 2.

This is possible when $\sin^2x=1/k$

Therefore $\cos^2x=1-\frac{1}{k}$

Now solving the asked expression we get

$1+\frac{5}{k} \sqrt{k-1}$

Answer 2

Answer:

as k sin2x + (1/k) cosec 2 x = 2

which is (ksin2x - 1)2 = 0

or sin2x = 1/k , cos2x = 1 -(1/k)

for cos2x + 5 sinx cosx + 6 sin 2x

= 5 - 5/k +6/k + 5/k*(k-1)1/2

= ( 1+5k+ 5*(k-1)1/2) / k

ans: none of the these