1. Trig Review Help

1.
Simplify the following expression in terms of x
sq(4tan^2x+4)

I did:
sq(4(tan^2x+1))
sq(4sec^2x)
=+/- 2secx

The correct answer is 2abs(secx) and I don't understand why, so I'd appreciate it if someone would explain it to me.

2.
Write tan(arc cos(x/3)) in algebraic form
I don't remember how to do this at all, so detailed explanation on what to do would be extremely helpful.

Any help is greatly appreciated.

2. Hello Naples
Originally Posted by Naples
1.
Simplify the following expression in terms of x
sq(4tan^2x+4)

I did:
sq(4(tan^2x+1))
sq(4sec^2x)
=+/- 2secx

The correct answer is 2abs(secx) and I don't understand why, so I'd appreciate it if someone would explain it to me.
You've done all the hard work! It's simply that the original expression, $\sqrt{4\tan^2x+4}$, is written as a positive square root. Hence it becomes the positive square root of $4\sec^2x$, which is written as an absolute value: $2|\sec x|$, so that when $x$ would make $\sec x$ negative, it still gives the corresponding positive value.

2.
Write tan(arc cos(x/3)) in algebraic form
I don't remember how to do this at all, so detailed explanation on what to do would be extremely helpful.

Any help is greatly appreciated.
Let $y = \arccos(x/3)$. Then we shall need to find $\tan y$.

$\cos y = \frac x3$

$\Rightarrow \sec y = \frac3x$

$\Rightarrow \sec^2 y = 1+\tan^2 y= \frac{9}{x^2}$

$\Rightarrow \tan^2y= \frac{9}{x^2}-1=\frac{9-x^2}{x^2}$

$\Rightarrow \tan y = \tan(\arccos (x/3) = \sqrt{\frac{9-x^2}{x^2}}$