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Let us consider the equation cos^4x/a+si...

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^8x/b^3+cos^8x/a^3` is

A

`sin^4x/b=cos^4x/a`

B

`sinx/a=cosx/b`

C

`sin^4x/b^2=cos^4x/a^2`

D

`sin^2x/a=cos^2x/b`

Text Solution

Verified by Experts

The correct Answer is:
C
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)`
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Knowledge Check

  • 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

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

  • The number of solutions of the equation (3+cos x)^(2)=4-2sin^(8)x" in "[0, 9pi) is equal to

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    6
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  • If "cos" x +"cos"^2 x -1, then the value of "sin"^12 x+3 "sin"^8x+"sin" ^6x-1 is

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