If sectheta-tantheta=(a+1)/(a-1), then c...

If s`ectheta-tantheta=(a+1)/(a-1)`, then `costheta`=

A

`(a^2+1)/(a^2-1)`

B

`(a^2-1)/(a^2+1)`

C

`(2a^2)/(a^2+1)`

D

`(2a^2)/(a^2-1)`

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