Trigo Qn

• Aug 27th 2010, 02:55 AM
Nuts
Trigo 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 AM
mr fantastic
Quote:

Originally Posted by Nuts
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.

On the one hand, $(\sec (A) - \tan (A))(\sec (A) + \tan (A)) = \sec^2 (A) - \tan^2 (A) = 1$ (this follows from the basic Pythagorean Identity).

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

Therefore .....
• Aug 27th 2010, 03:23 AM
Quote:

Originally Posted by Nuts
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.

$\displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA$

We want to find

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

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

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

from which follows the result.
• Aug 27th 2010, 03:33 AM
Nuts
Quote:

Originally Posted by mr fantastic
On the one hand, $(\sec (A) - \tan (A))(\sec (A) + \tan (A)) = \sec^2 (A) - \tan^2 (A) = 1$ (this follows from the basic Pythagorean Identity).

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

Therefore .....

Hmm i do not understand how does Sec A - tan A = 2 changes to $(\sec (A) - \tan (A))(\sec (A) + \tan (A)) = 2 (\sec (A) + \tan (A))$ . Could you explain (Nerd)
• Aug 27th 2010, 03:38 AM
Prove 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

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

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

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

Now use the Pythagorean Identity $1 + \tan^2{\theta} \equiv \sec^2{\theta}$ to simplify the LHS and you should be able to solve for $\sec{A} + \tan{A}$.
• Aug 27th 2010, 03:56 AM
Nuts
Quote:

$\displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA$

We want to find

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

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

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

from which follows the result.

Ok i solved it thanks! :)
• Aug 27th 2010, 04:00 AM