# Thread: solving Natural Logs for x and a Trig identity question.

1. ## solving Natural Logs for x and a Trig identity question.

Hey.

Code:
2Ln2 - Ln(x-1) = ln(2x)
That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.

also, i cant seem to finish this off:
Code:
Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)
This is what i have so far:
Code:
-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
-> 1/(tan^2 θ - sin^2 θ)
-> (1/((cos^2 θ)/(sin^2 θ))) - (1/sin^2 θ)
-> ((sin^2 θ)/(cos^2 θ)) + ((sin^2 θ)/ 1)
i am stuck there... unless of course the actually DO add together to make
Code:
(sin^4 θ)/(cos^2 θ)
Any help would be greatly appreciated!

2. Originally Posted by squrrilslayer
Hey.

Code:
2Ln2 - Ln(x-1) = ln(2x)
That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.

b ln(a) = ln(a^b)

Thus

ln(4) - ln(x-1) = ln(2x)

But
ln(a) - ln(b) =ln(a/b)

Thus

ln(4/(x-1)) = ln(2x)

Thus

4/x-1 = 2x

Hence
2x^2 - 2x = 4

x^2 - x - 2 = 0

x^2 - 2x + x -2 = 0

x(x-2) + (x-2) = 0

(x+1)(x-2)=0

x = -1 or x= 2

But since ln(-ve) is not defined thus x = 2

3. Originally Posted by squrrilslayer

Code:
Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)
This is what i have so far:

Code:
-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
-> tan^2θ - sin^2 θ
-> sin^2 θ /cos^2 θ - sin^2 θ
-> sin^2θ( 1/cos^2 θ -1)
-> sin^2 θ ( sec^2 θ -1)
-> sin^2θ (tan^2 θ)
-> sin^2 θ x sin^2 θ/cos^2 θ
-> sin^4 θ/cos^2 θ
I have corrected in code

4. wow thanks that was quick. I see now on the first one how you did that
as for the second, you got the answer so im reading through step by step analyzing what you did.

Thanks again!
[resolved]

5. ## sorry re on trig idents

i still dont understand what's been done:

-> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
-> tan^2θ - sin^2 θ
-> sin^2 θ /cos^2 θ - sin^2 θ <<here
-> sin^2θ( 1/cos^2 θ -1) <<here
-> sin^2 θ ( sec^2 θ -1) <<here
-> sin^2θ (tan^2 θ) <<here
-> sin^2 θ x sin^2 θ/cos^2 θ
-> sin^4 θ/cos^2 θ

like how did

sin^2 θ /cos^2 θ - sin^2 θ

become:

sin^2θ( 1/cos^2 θ -1) ?

and so on so forth?

Thanks if someone could clear this up!

6. Line (3) is (effectively) "x/y - x".
Line (4) is (effectively) "x(1/y - 1)".
The author factored the "x" out front.

To go from Line (5) to Line (6), apply one of the trig identities they gave you.

7. $\displaystyle tan^2(\theta) - sin^2(\theta) = \frac{sin^2(\theta)}{cos^2(\theta)} - sin^2 (\theta)$

$\displaystyle \frac{sin^2(\theta)}{cos^2(\theta)} - sin^2(\theta)$

$\displaystyle = sin^2(\theta)\times \frac{1}{cos^2(\theta)} - sin^2(\theta) \times 1$

Take $\displaystyle sin^2(\theta)$ common

$\displaystyle = sin^2(\theta) (\frac{1}{cos^2(\theta)} -1)$-------(1)

Using the rule that

sec(x) = 1/cos(x)

Replace $\displaystyle 1/cos^2(\theta)$ by $\displaystyle sec^2(\theta)$

= sin^2 θ ( sec^2 θ -1)......................(2)

Using the property that

sec^2 (x) = 1 + tan^2 (x)

Replace $\displaystyle sec^2(\theta)$ by $\displaystyle tan^2(\theta) - 1$

which gives

$\displaystyle = sin^2(\theta)\times (tan^2(\theta))$

$\displaystyle \frac{sin^2 \theta \times sin^2\theta}{cos^2 \theta}$

-----------------------------

Tell the step you fail to understand now