If `cos25^0+sin25^0=p ,` then `cos50^0` is

Options

A. `sqrt(2-p^2)`
B. `-sqrt(2-p^2)`
C. `psqrt(2-p^2)`
D. `-psqrt(2-p^2)`

Explanation

`cos25^@+sin25^@ = p`

Squaring both sides,

`=>cos^2 25^@+sin^2 25^@+2sin25^@cos25^@ = p^2`

`=>1+sin 50^@ = p^2`

`=>sin50^@ = p^2-1`

`=>cos50^@ = sqrt(1-sin^2 50^@)`

`=sqrt(1-(p^4+1-2p^2))`

`=psqrt(2-p^2)`

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