If `sin^(2)x+sin^(2)y=1`, then what is the value of `cot(x+y)`?

Options

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

Answer

C

Explanation

Given `sin^(2)x+sin^(2)y=1`
`rArrsin^(2)x=1-sin^(2)y`
`rArrsin^(2)x=cos^(2)y`
`rArrsinx=cosy`
Similarly By considering
`sin^(2)y=1-sin^(2)x`, we have
cos x = sin y Now , Consider cot `(x+y)=(cos(x+y))/(sin(x+y))`
`=(cosxcosy-sinxsiny)/(sin(x+y))`
`=(cosxcosy-cosxcosy)/(sin(x+y))=0`