Math, asked by sumatisrivastav, 1 year ago

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

sumatisrivastav: i had given the exam but question paper was not given along with it but what i still remember is something lyk x+1/x

QGP: Does it have any option like this?

sumatisrivastav: x-1/x

sumatisrivastav: i didn't remember last two options but they were having denominator as x^2

QGP: Let me recheck my answer again

sumatisrivastav: okk

QGP: Since I have not found any error in my approach, I am answering the question. It's going to take a fairly long time to type

QGP: About 20 minutes

QGP: Please wait till that

sumatisrivastav: ok i m waitng

Answers

Answered by ayushm

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

sumatisrivastav: ans should only be in term of y not any other trignometric ratio

ayushm: Edited. Check again.

sumatisrivastav: options given for this answers were not in under root

ayushm: 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?

Answered by QGP

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.

QGP: Hope it helps

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QGP: I hope you understand it

sumatisrivastav: yes thank u...i will write it down to undrstnd it more clearly

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