If `sec theta + tan theta= p`, then `costheta` is

Options

A. `(rho^(2) + 1)/( rho^(2) - 1)`
B. ` ( rho^(2) - 1)/(( rho^(2)+ 1)^(2))`
C. `(2 rho)/( rho^(2) + 1)`
D. ` ( 4 rho^(2))/(( rho^(2) + 1) ^(2))`

Answer

C

Explanation

`sec theta + tan theta = rho rArr (1)/( cos theta ) + (sin theta)/( cos theta) = rho`
or ` 1+ sin theta = rho cos theta " "` . . . (i)
On squaring both sides, we get
`1 + sin ^(2) theta + 2 sin theta = rho^(2) cos^(2) theta `
`rArr 1 + 1 - cos ^(2) theta + 2 sin theta = rho^(2) cos^(2) theta `
`rArr 2 + 2 sin theta - cos ^(2) theta = rho^(2) cos^(2) theta `
`rArr 2 ( 1 + sin theta) - cos^(2) theta = rho^(2) cos ^(2) theta `
`rArr 2 xx rho cos theta - cos^(2) theta = rho^(2) cos^(2) theta " "` [from Eq. (i) ]
`rArr 2 rho cos theta = rho^(2) cos^(2) theta + cos^(2) theta `
`rArr 2 rho cos theta = cos ^(2) theta (1 + rho^(2))`
`:. cos^(2) theta = (2 rho cos theta )/( (1 + rho ^(2))) rArr cos theta = (2 rho )/( 1 + rho^(2))`