Problem Help

**gabrie30** · June 3rd 2008, 11:26 AM

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!

**Jhevon** · June 3rd 2008, 11:49 AM

Originally Posted by gabrie30

Having trouble with this problem....

sec^2-csc^2=csc^2(sec^2-2) "after ^2 there is theta"

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

csc(tan^-1(-.5)

hummm,

Thanks if anyone can help!

what are you supposed to be doing here...?

**Moo** · June 3rd 2008, 11:53 AM

Originally Posted by Jhevon

what are you supposed to be doing here...?

Just moo

@ gabrie30 : can you put parenthesis ? and use t for theta

**gabrie30** · June 3rd 2008, 06:52 PM

Originally Posted by Jhevon

what are you supposed to be doing here...?

Finding the exact value for the inverse tan and "establish the identity" for the other

**Gusbob** · June 4th 2008, 04:09 AM

Originally Posted by gabrie30

Establish the Identity:
(sec^2t)-(csc^2t)=csc^2t(sec^2t-2)

$LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t}$
$= \frac{sin^2t- cos^2t}{cos^2t\,sin^2t}$
$= \frac{(1-cos^2t) - cos^2t}{cos^2t\,sin^2t}$
$= \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}$
$= \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)$
$= csc^2t ( sec^2t -2)$

Find the exact Value:
$csc(tan^{-1}(-.5))$

Note that arctan(-.5) is an angle
First let $y = arctan (-.5)$
$tan y = -.5$
If you recall basic trigonometric ratios, tangent = $\frac{opposite\,side}{adjacent\,side} = - \frac{1}{2}$ in this case.
Now we know that the opposite side is $\pm 1$ when the adjacent side is $\mp 2$
Using Pythagoras's theorem, we can deduce that the hypotenuse is $\sqrt{5}$
$csc \,y = \frac{hypotenuse}{opposite} = \pm \frac{\sqrt5}{1}$
= $\pm \sqrt5$

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

**gabrie30** · June 5th 2008, 01:06 PM

[quote=Gusbob;153603] $LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t}$
$= \frac{sin^2t- cos^2t}{cos^2t\,sin^2t}$
$= \frac{(1-cos^2t) - cos^2t}{cos^2t\,sin^2t}$
$= \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}$

Can someone make this step more clear.

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

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