If `cos^(2)alpha+cos^(2)beta=2`, then the value of `tan^(3)alpha+sin^(5)beta` is

Options

A. `-1`
B. `0`
C. `1`
D. `(1)/(sqrt(3))`

Answer

b

Explanation

`cos^(2)alpha+cosbeta=2`
Put value of `alpha=beta=0^(@)`
`rArrcos^(2)0^(@)+cos^(2)0^(@)=2`
`rArr(1)^(2)+(1)^(2)=2`
`rArr2=2` [it satisfies the equation]
`=tan^(2)alpha+sin^(3)beta`
`=tan^(2)alpha+sin^(2)beta`
`=tan^(2)0^(@)+sin^(2)0^(@)`
`=0+0=0`