Let us consider the equation `cos^4x/a+sin^4x/b=1/(a+b),x in[0,pi/2],a,bgt0`
The value of `sin^2x` in terms of a and b is

Options

A. `sqrt(ab)`
B. `b/(a+b)`
C. `(b^2-a^2)/(a^2+b^2)`
D. `(a^2+b^2)/(b^2-a^2)`

Answer

B

Explanation

We have, `cos^4x/a+sin^4x/b=1/(a+b)=(cos^2x+sin^2x)/(a+b)`
`rArr cos^2x(cos^2x/a-1/(a+b))=sin^2x(1/(a+b)-sin^2x/b)`
`rArrcos^2x((bcos^2x-asin^2x)/(a(a+b)))=sin^2x((bcos^2x-asin^2x)/(b(a+b)))`
`rArrcos^2x/a=sin^2x/b`
`rArr(1-sin^2x)/a=sin^2x/b`
`rArr sin^2x=b/(a+b)and cos^2x=a/(a+b)`
`:. sin^8x/b^3+cos^8x/a^3=b^3/(b^3(a+b)^4)+a^4/(a^3(a+b)^4)=1/((a+b)^3)`