1.

For `0ltphiltpi//2," if" x=sum_(n=0)^(oo) cos^(2n)phi,y=sum_(n=0)^(oo) sin^(2m)phi,z=sum_(n=0)^(oo)cos^(2n)phisin^(2n)phi`,thenA. `xyz=xz+y`B. `xyz=xy+z`C. `xyz=x+y+z`D. `xyz=yz+x`

Answer» Correct Answer - B::C
All are infinte geometric progression with common ratio lt 1
`x=1/(1-cos^2phi)=1/sin^2phi,y=1/(1-sin^2phi)=1/cos^2phi`,
`z=1/(1-cos^2phisin^2phi)`
Now, `xy+z=1/(sin^2phicos^2phi)+1/(1-sin^2phicos^2phi)`
`=1/(sin^2phicos^2phi(1-sin^2phicos^2phi))`
`or xy+z=xyz ...(i)`
Clearly, `x+y=(sin^2phi+cos^2phi)/(sin^2phicos^2phi)=xy`
`:. x+y+z=xyz` [using Eq. (i)]


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