Prove : (1+cosA - sinA)/(1+cosA + sinA) = secA - tanA

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The discussion focuses on proving the trigonometric identity (1+cosA - sinA)/(1+cosA + sinA) = secA - tanA using fundamental trigonometric identities. Participants express confusion about the correct approach, with suggestions to manipulate the left-hand side (LHS) by multiplying by a specific factor to simplify the expression. The conversation highlights the importance of factoring both the numerator and denominator to facilitate cancellation of terms. Ultimately, one participant successfully factors the numerator, confirming that it simplifies to match the right-hand side (RHS) of the equation. The thread concludes with a sense of accomplishment as the proof is verified.
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Homework Statement


Prove that
(1+cosA - sinA)/(1+cosA + sinA) = secA - tanA



Homework Equations


sin^2A + cos^2A = 1
tanA = sinA/cosA
cotA = cosA/sinA
1 + cot^2A = cosec^2A
tan^2A + 1 = sec^2A
cosecA = 1/sinA
secA = 1/cosA
cotA = 1/tanA
(Only use the above identities to prove the question)

The Attempt at a Solution


I'm stumped at this question. I have attempted various methods using the formulas that I know(stated above)and also trying to work on both sides but to no avail. I understand that by cross multiplying we can easily prove it but the correct way seems to just be by making either the LHS or RHS equal to the other,respectively. Can anyone help?
 
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Multiply the LHS by:
<br /> \frac{1+\cos A-\sin A}{1+\cos A-\sin A}<br />
Expand.
 
hunt_mat said:
Multiply the LHS by:
<br /> \frac{1+\cos A-\sin A}{1+\cos A-\sin A}<br />
Expand.

Don't i need to account for the RHS also? Or are we rationalizing like we do for surds?
 
You're multiplying by 1, so you only need to do this for the LHS, expand ans you'll see that things cancel and you end up with the RHS
 
hunt_mat said:
You're multiplying by 1, so you only need to do this for the LHS, expand ans you'll see that things cancel and you end up with the RHS

I think i went wrong?

I finalized to ,
2+2cosA - 2sinA - 2sinAcosA
----------------------------
1 + 2cosA + cos^2A - sin^2A

Sorry if this is hard to read,i don't know how to use latex. :/
 
You're perfectly correct, you write 1=sin^{2}A+\cos^{2}A in the deominator, does the numorator factor (hint, it does).

Mat
 
hunt_mat said:
You're perfectly correct, you write 1=sin^{2}A+\cos^{2}A in the deominator, does the numorator factor (hint, it does).

Mat

Do you group the sin and the cos together before factoring? If so,where do we put the troublesome sinAcosA?

i'm really bad at this. I only managed to factor the denominator to cosA(2+2cosA)
 
You're halfway there! Look for the factor (2-2cosA) in the numorator, and then they should cancel.

Mat
 
Okay wait i cheated a little by looking at my RHS that i have converted into a fraction and i got it. Thanks alot! the numerator factors into ( 1-sinA) ( 2+2cosA) am i right? :)
 
  • #10
Well done. You've done it.
 
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