# need help for test monday!!!

#### sam

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

#### CaptainBlack

MHF Hall of Fame
sam said:
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

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

RonL

#### sam

trig

that's the way the prob is in my textbook

#### sam

it says to simplify the trigonometric expression

#### sam

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

#### CaptainBlack

MHF Hall of Fame
sam said:
can't do trig! stuff like sec^2(x)-1/sec^2(x).
Help me!
more to come
Simplify:

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

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

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

If that is simpler.

RonL

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#### c_323_h

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?

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#### c_323_h

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.

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#### 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.

#### CaptainBlack

MHF Hall of Fame
sam said:
also 1+cos(y)/1+sec(y)
This is ambiguous, it could mean:

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

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

$$\displaystyle (3)\ \ 1+\frac{\cos(y)}{1}+\sec(y)$$ (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