Hi i need help for this question. Stucked at it for a long time
Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without
solving for A.
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 .....
$\displaystyle \displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA$
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.
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 $\displaystyle \sec{A} + \tan{A}$.