Hi i need help for this question. Stucked at it for a long time (Doh)

Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without

solving for A.

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- Aug 27th 2010, 02:55 AMNutsTrigo Qn
Hi i need help for this question. Stucked at it for a long time (Doh)

Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without

solving for A. - Aug 27th 2010, 03:07 AMmr fantastic
On the one hand, $\displaystyle (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = \sec^2 (A) - \tan^2 (A) = 1$ (this follows from the basic Pythagorean Identity).

On the other hand, it's given that $\displaystyle (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = 2 (\sec (A) + \tan (A))$.

Therefore ..... - Aug 27th 2010, 03:23 AMArchie Meade
$\displaystyle \displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA$

We want to find

$\displaystyle \displaystyle\frac{1+SinA}{CosA}$

$\displaystyle \displaystyle\ (1-SinA)(1+SinA)=2CosA(1+SinA)$

$\displaystyle 1-Sin^2A=Cos^2A=2CosA(1+SinA)\Rightarrow\ CosA=2(1+SinA)$

from which follows the result. - Aug 27th 2010, 03:33 AMNuts
- Aug 27th 2010, 03:38 AMProve It
Usually the easiest way to simplify trigonometric expressions is to multiply by the conjugate, then apply a trigonometric identity.

In your case, you are told

$\displaystyle \sec{A} - \tan{A} = 2$

$\displaystyle (\sec{A} - \tan{A})(\sec{A} + \tan{A}) = 2(\sec{A} + \tan{A})$

$\displaystyle \sec^2{A} - \tan^2{A} = 2(\sec{A} + \tan{A})$.

Now use the Pythagorean Identity $\displaystyle 1 + \tan^2{\theta} \equiv \sec^2{\theta}$ to simplify the LHS and you should be able to solve for $\displaystyle \sec{A} + \tan{A}$. - Aug 27th 2010, 03:56 AMNuts
- Aug 27th 2010, 04:00 AMArchie Meade