Having trouble with this problem....

Establish the Identity:

(sec^2t)-(csc^2t)=csc^2t(sec^2t-2)

and this ones seems easy but im not sure...

Find the exact Value:

csc(tan^-1)(-.5)

hummm,

Thanks if anyone can help!

Results 1 to 6 of 6

- Jun 3rd 2008, 11:26 AM #1

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## Problem Help

Having trouble with this problem....

Establish the Identity:

(sec^2t)-(csc^2t)=csc^2t(sec^2t-2)

and this ones seems easy but im not sure...

Find the exact Value:

csc(tan^-1)(-.5)

hummm,

Thanks if anyone can help!

- Jun 3rd 2008, 11:49 AM #2

- Jun 3rd 2008, 11:53 AM #3

- Jun 3rd 2008, 06:52 PM #4

- Joined
- May 2008
- Posts
- 10

- Jun 4th 2008, 04:09 AM #5

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$\displaystyle LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t} $

$\displaystyle = \frac{sin^2t- cos^2t}{cos^2t\,sin^2t} $

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

$\displaystyle = \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}$

$\displaystyle = \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)$

$\displaystyle = csc^2t ( sec^2t -2)$

Find the exact Value:

$\displaystyle csc(tan^{-1}(-.5))$

First let $\displaystyle y = arctan (-.5) $

$\displaystyle tan y = -.5 $

If you recall basic trigonometric ratios, tangent = $\displaystyle \frac{opposite\,side}{adjacent\,side} = - \frac{1}{2}$ in this case.

Now we know that the opposite side is $\displaystyle \pm 1 $ when the adjacent side is $\displaystyle \mp 2$

Using Pythagoras's theorem, we can deduce that the hypotenuse is $\displaystyle \sqrt{5}$

$\displaystyle csc \,y = \frac{hypotenuse}{opposite} = \pm \frac{\sqrt5}{1}$

= $\displaystyle \pm \sqrt5$

You might want to draw a triangle if it helps you to visualise

- Jun 5th 2008, 01:06 PM #6

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- May 2008
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[quote=Gusbob;153603]$\displaystyle LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t} $

$\displaystyle = \frac{sin^2t- cos^2t}{cos^2t\,sin^2t} $

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

**$\displaystyle = \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}$**

Can someone make this step more clear.

$\displaystyle = \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)$

$\displaystyle = csc^2t ( sec^2t -2)$