1. ## identity

i don't know how to do this identity:

tan ^4-sec^4=...

can you help me? i hope not to disturb you..but i'm reviewing 2 chapters made so far...and i truly need your help.

2. Hello, alessandromangione!

You don't have an identity.
The best we can do is simplify the expression.
. . (And how far should we go?)

Simplify:. . $\displaystyle \tan^4\!x -\sec^4\!x$

$\displaystyle \tan^4\!x - \sec^4\!x \;=\; -(\sec^4\!x - \tan^2\!x)$

. . . . . . . . .$\displaystyle =\;-\underbrace{(\sec^2\!x-\tan^2\!x)}_{\text{This is 1}}(\sec^2\!x + \tan^2\!x)$

. . . . . . . . .$\displaystyle =\; -(\sec^2\!x + \tan^2\!x)$

. . . . . . . . .$\displaystyle =\;\begin{Bmatrix}-(\tan^2\!x + 1 + \tan^2\!x) &=& -2\tan^2\!x - 1 \\ \\ -(\sec^2\!x + \sec^2\!x - 1) &=& -2\sec^2\!x + 1 \\ \\ -\left(\dfrac{1}{\cos^2\!x} + \dfrac{\sin^2\!x}{\cos^2\!x}\right) &=& -\dfrac{1 + \sin^2\!x}{\cos^2\!x} \end{Bmatrix}$

3. i would ike that u help me on doing these ones too...

sec^4 - 2 sec^2tan^2+ tan^4=

sec sin / tan -1

use a right triangle to write the expression as an algebraic expression. Assume that v is positive and in the domainof the given inverse trigonometric function.

8)cos ( inverse sin v )

9) sin (inverse sec square root of v^2+9/v) . The possible choices are :

a)v square root of v^2 +3/v^2+3
b) v suqare root of v^2/2
c) 3 square root of v^2 + 9 / v^2 +9
d) square root of v^2 +3/v^2+3

find the exact value of
inverse cos ( cos (-3 radian / 5) )

i have to do a test to catch up and raise my grades...really that you guys help me!

4. Originally Posted by Soroban
Hello, alessandromangione!

You don't have an identity.
The best we can do is simplify the expression.
. . (And how far should we go?)

$\displaystyle \tan^4\!x - \sec^4\!x \;=\; -(\sec^4\!x - \tan^2\!x)$

. . . . . . . . .$\displaystyle =\;-\underbrace{(\sec^2\!x-\tan^2\!x)}_{\text{This is 1}}(\sec^2\!x + \tan^2\!x)$

. . . . . . . . .$\displaystyle =\; -(\sec^2\!x + \tan^2\!x)$

. . . . . . . . .$\displaystyle =\;\begin{Bmatrix}-(\tan^2\!x + 1 + \tan^2\!x) &=& -2\tan^2\!x - 1 \\ \\ -(\sec^2\!x + \sec^2\!x - 1) &=& -2\sec^2\!x + 1 \\ \\ -\left(\dfrac{1}{\cos^2\!x} + \dfrac{\sin^2\!x}{\cos^2\!x}\right) &=& -\dfrac{1 + \sin^2\!x}{\cos^2\!x} \end{Bmatrix}$

thank you dude!!