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# How do I prove that (cosθ/(1+sinθ ))+((1+sinθ)/cosθ) =2 secθ.?

Can some one please help walk me through this? Yes, its a homework question but I am stuck and really would appreciate help here.

### 5 Answers

- Wayne DeguManLv 71 month agoFavorite Answer
cosθ/(1 + sinθ) + (1 + sinθ)/cosθ = 2

Working on the left side and forming a common denominator we have:

[cos²θ + (1 + sinθ)²]/cosθ(1 + sinθ)

i.e. [cos²θ + 1 + 2sinθ + sin²θ]/cosθ(1 + sinθ)

Now, as sin²θ + cos²θ = 1 we have:

(2 + 2sinθ)/cosθ(1 + sinθ)

i.e. 2(1 + sinθ)/cosθ(1 + sinθ)

so, 2/cosθ

Hence, 2secθ => R.H.S.

:)>

- ?Lv 71 month ago
let c=cos(θ), s=sin(θ)

LHS = c/(1+s) + (1+s)/c

// use the Pythagorean identity

s² + c² = 1, thus 1-s²=c²

// multiply 1+s by its conjugate

// difference of two squares

(1+s)(1-s) = 1-s² = c²

// multiply first term by 1 = (1-s)/(1-s)

c/(1+s) = c(1-s)/(1-s²) = c(1-s)/c²

= (1-s)/c

LHS = (1-s)/c + (1+s)/c = 2/c

// sec(θ) = 1/cos(θ)

LHS = 2 sec(θ) = RHS

While this is not so straight forward as other answers, it illustrates three key ideas in proving trig identities:

use algebra: common denominator, factoring, multiply denominator by conjugate, difference of two squares

use definitions for trig functions, mostly in terms of sine and cosine

use well known trig identities: the Pythagorean identity shows up over and over again.

Last, I like substitution, particularly when posting on line. It can reduce the clutter of symbols and make the underlying structure more obvious.

- la consoleLv 71 month ago
= [cos(θ)/{1 + sin(θ)}] + [{1 + sin(θ)}/cos(θ)]

= [cos²(θ) + {1 + sin(θ)}.{1 + sin(θ)}] / [{1 + sin(θ)}.cos(θ)]

= [cos²(θ) + 1 + sin(θ) + sin(θ) + sin²(θ)] / [{1 + sin(θ)}.cos(θ)] → recall: cos²(θ) + sin²(θ) = 1

= [2 + 2.sin(θ)] / [{1 + sin(θ)}.cos(θ)]

= 2.[1 + sin(θ)] / [{1 + sin(θ)}.cos(θ)]

= 2/cos(θ) → recall: sec(x) = 1/cos(x)

= 2.sec(θ)

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- lenpol7Lv 71 month ago
(cosθ/(1+sinθ ))+((1+sinθ)/cosθ) =2 secθ.=>

NB Addition of fractions. [Cos^2 + ( 1 + Sin)^2] / [Cos(1 + Sin)] =>[Cos^2 + 1 + 2Sin + Sin^2] / (Cos(1 + Sin))[1 + 1 + 2Sin] / (Cos + (1 + Sin) [2 + 2Sin] / Cos(1 + Sin) => 2(1 + Sin)] / Cos(1 + Sin)(1 + Sin) Cancels down Hence 2 /Cos => 1 x 1/Cos => 2Sec (X) NB I have omitted 'X' / 'Theta' for clarity .