# need help for test monday!!!

• April 29th 2006, 09:32 AM
sam
need help for test monday!!!
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
• April 29th 2006, 09:52 AM
CaptainBlack
Quote:

Originally Posted by sam
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come

This is not a question, what do you want to do with

$
\sec^2(x)-1/\sec^2(x)?
$

RonL
• April 29th 2006, 10:45 AM
sam
trig
that's the way the prob is in my textbook
• April 29th 2006, 10:54 AM
sam
it says to simplify the trigonometric expression
• April 29th 2006, 10:54 AM
sam
also 1+cos(y)/1+sec(y)
• April 29th 2006, 11:22 AM
CaptainBlack
Quote:

Originally Posted by sam
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come

Simplify:

$
\sec^2(x)-1/\sec^2(x)=1/\cos^2(x)-\cos^2(x)=\frac{1-\cos^4(x)}{\cos^2(x)}
$

$
=\frac{(1-\cos^2(x))(1+\cos^2(x))}{\cos^2(x)}=\frac{\sin^2(x )(1+\cos^2(x))}{\cos^2(x)}
$

$
=\tan^2(x)(1+\cos^2(x))
$

If that is simpler.

RonL
• April 29th 2006, 02:41 PM
c_323_h
is it $\frac{\sec^2x-1}{\sec^2x}$ or
$\sec^2x-\frac{1}{sec^2x}$

this is what i got for the first one

$\frac{\sec^2x-1}{\sec^2x}$
$\frac{\sec^2x-1}{1} \bullet \frac{1}{sec^2x}$
since
$\frac{1}{secx}=cosx$ you can square both sides to get
$\frac{1}{sec^2x}=cos^2x$
substitute this in to get
$(sec^2x-1)(cos^2x)$=
$(tan^2x)(cos^2x)$

can someone tell me if that dot is the correct multiplication operator on the second line?
• April 29th 2006, 03:08 PM
c_323_h
and for the second:

$
\sec^2x-\frac{1}{sec^2x}
$

since
$\frac{1}{\sec}=\cos$ we have,
$\sec^2x-\cos^2x$

there are trig identities you must understand and know how to manipulate them.
• April 29th 2006, 05:57 PM
c_323_h
by the way, there are no unique answers, there are many different correct answers. CaptainBlack's or my answers are just fine.
• April 29th 2006, 10:41 PM
CaptainBlack
Quote:

Originally Posted by sam
also 1+cos(y)/1+sec(y)

This is ambiguous, it could mean:

$
(1)\ \ \frac{1+\cos(y)}{1+\sec(y)}
$

$
(2)\ \ 1+\frac{\cos(y)}{1+\sec(y)}
$

$
(3)\ \ 1+\frac{\cos(y)}{1}+\sec(y)
$
(not too likely)

Now the first of these seems the most likely, but you are leaving us guessing.

If you are writing this using plain ASCII then write it as:

(1+cos(y))/(1+sec(y))

The rule is: when something is ambiguous when written in plain ASCII then
add brackets until only the intended meaning is possible.

RonL