1.

`tan^(-1)[2cos(2sin^(-1)""(1)/(2))]`

Answer» `tan^(-1)[2cos(2sin^(-1)""(1)/(2))]`
`=tan^(-1)[2cos(2sin^(-1)sin""(pi)/(6))]`
`=tan^(-1)[2cos(2.(pi)/(6))]`
`=tan^(-1)[2.cos""(pi)/(3)]`
`=tan^(-1)(2.(1)/(2))=tan^(-1)(1)`
`tan^(-1)(tan""(pi)/(4))=(pi)/(4)`


Discussion

No Comment Found