Answers

2016-04-04T23:07:33+05:30
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
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ans should only be in term of y not any other trignometric ratio
Edited. Check again.
options given for this answers were not in under root
I'm not aware of what options you're referring to. I believe my answer isn't wrong. Can you find any error in it?
The Brainliest Answer!
2016-04-05T00:11:27+05:30
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.
2 5 2
yes thank u...i will write it down to undrstnd it more clearly
Thanks for marking it as the brainliest.
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