Question

if sec x + tan x = y then cos=? in the terms of y

Answer 1

$secx + tanx = y\\\\sqrt{tan x^{2} + 1} + tanx = y\\ \\ Square both sides,\\\\ tan^2 x + 1 + tan^2x + 2tanx \sqrt{tan^2x + 1} = y^2\\\\ 2tan^2x + 2tanx \sqrt{tan^2x + 1} = y^2\\\\ 2tanx(tanx + \sqrt{tan^2x + 1} ) = y^2\\\\ 2ytanx = y^2\\\\ 2tanx = y\\\\ tanx = y/2\\ \\Therefore, \\\\ cosx = 2/(\sqrt{y^2-4]})$Q.E.D

Answer 2

I am answering your original question:

If sec θ + tan θ = x   then   cos θ=?

SOLUTION:

sec θ + tan θ = x
∴1/cosθ  +  sinθ/cosθ  =  x
∴(1+sinθ)/cosθ = x
∴1+sinθ = x cosθ   ----------- (1)

Now, we know that

sin²θ + cos²θ = 1
∴cos²θ = 1 - sin²θ
∴cos²θ = (1+sinθ)(1-sinθ)          [∵a² - b² = (a+b)(a-b)  ] 
∴cos²θ = x cosθ (1-sinθ)            [From (1) ]
∴cos²θ/x cosθ  =  1-sinθ
∴cosθ/x = 1 - sinθ
∴sinθ = 1 - cosθ/x ----------------------- (2)

From (1), we know
1 + sinθ = x cosθ
∴1 + 1 - cosθ/x  =  x cosθ           [From (2) ]
∴2 - cosθ/x = x cosθ
∴(2x - cosθ) / x   =  x cosθ
∴ 2x - cos θ  =  x² cosθ
∴2x = x² cosθ + cosθ
∴2x = cosθ (x²+1)
∴2x / (x²+1) = cosθ

∴ cos θ = 2x/(x²+1)

Please note that I have used brackets for denoting common denominators. Where I have not used brackets, it means the denominator is for a single term only.