identity

**alessandromangione** · December 13th 2009, 01:43 PM

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.

**Soroban** · December 13th 2009, 02:01 PM

Hello, alessandromangione!

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

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

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

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

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

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

**alessandromangione** · December 13th 2009, 02:08 PM

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!

**alessandromangione** · December 13th 2009, 02:09 PM

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?)

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

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

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

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

thank you dude!!

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