can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
Simplify:Originally Posted by sam
$\displaystyle
\sec^2(x)-1/\sec^2(x)=1/\cos^2(x)-\cos^2(x)=\frac{1-\cos^4(x)}{\cos^2(x)}
$
$\displaystyle
=\frac{(1-\cos^2(x))(1+\cos^2(x))}{\cos^2(x)}=\frac{\sin^2(x )(1+\cos^2(x))}{\cos^2(x)}
$
$\displaystyle
=\tan^2(x)(1+\cos^2(x))
$
If that is simpler.
RonL
is it $\displaystyle \frac{\sec^2x-1}{\sec^2x}$ or
$\displaystyle \sec^2x-\frac{1}{sec^2x}$
this is what i got for the first one
$\displaystyle \frac{\sec^2x-1}{\sec^2x}$
$\displaystyle \frac{\sec^2x-1}{1} \bullet \frac{1}{sec^2x}$
since
$\displaystyle \frac{1}{secx}=cosx$ you can square both sides to get
$\displaystyle \frac{1}{sec^2x}=cos^2x$
substitute this in to get
$\displaystyle (sec^2x-1)(cos^2x)$=
$\displaystyle (tan^2x)(cos^2x)$
can someone tell me if that dot is the correct multiplication operator on the second line?
and for the second:
$\displaystyle
\sec^2x-\frac{1}{sec^2x}
$
since
$\displaystyle \frac{1}{\sec}=\cos$ we have,
$\displaystyle \sec^2x-\cos^2x$
there are trig identities you must understand and know how to manipulate them.
This is ambiguous, it could mean:Originally Posted by sam
$\displaystyle
(1)\ \ \frac{1+\cos(y)}{1+\sec(y)}
$
$\displaystyle
(2)\ \ 1+\frac{\cos(y)}{1+\sec(y)}
$
$\displaystyle
(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