1. ## 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!

2. 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...?

3. Originally Posted by Jhevon
what are you supposed to be doing here...?
Just moo

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

4. 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

5. Originally Posted by gabrie30
Establish the Identity:
(sec^2t)-(csc^2t)=csc^2t(sec^2t-2)
$\displaystyle LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t}$
$\displaystyle = \frac{sin^2t- cos^2t}{cos^2t\,sin^2t}$
$\displaystyle = \frac{(1-cos^2t) - cos^2t}{cos^2t\,sin^2t}$
$\displaystyle = \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}$
$\displaystyle = \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)$
$\displaystyle = csc^2t ( sec^2t -2)$

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

Note that arctan(-.5) is an angle
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

6. [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.

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

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