Please help me with these identity proofs!!

Question

1) (cscx + cotx) / (tanx + sinx) = cotxcscx

 

2) (1/secx) -secx = -sinx

 

3) (1-tan^2x / 1-cot^2x) = -tan^2x

Answer 1

Hi Olivia,

 

Here are some suggestions on the first one. We used to say back in trig class when you have to prove an identity and you do not know how to proceed, then PUNT. That was simply a code word for "convert to sine and cosine".

 

So, let's work on the left side and PUNT.

 

1/sinx + cosx/sinx

----------------------

sinx/cosx + sinx

 

To simplify this "complex" fraction multiply every term in the numerator and denominator by sinxcosx (the LCD of all the denominators).

 

That gives:

 

cosx+cos²x

----------------------

sin²x + sin²xcosx

 

In the numerator factor out a cosx; in the denominator factor out a sin²x.

 

cosx(1+cosx)

-----------------

sin²x(1+cosx)

 

The factor (1+cosx) cancels leaving:

 

cosx     1

------- ------

sinx      sinx

 

Notice how I split the sin²x into two factors.

 

That converts to cotxcscx.

 

Q.E.D.

 

Is there another way to prove this identity? Probably. But PUNTING gets you there too.

Answer 2

Whenever I deal with trig identities I find it really useful to change everything in the equation to some form of sine and/or cosine. Then pretty much every trig identity problem simply becomes an exercise of algebra, which requires no memorization. In trig, limiting memorization is always a good thing.

 

1) Let's see how the terms translate:

 

csc = 1/sin

cot = cos/sin

tan = sin/cos

 

(1/sin(x) + cos(x)/sin(x)) / (sin(x)/cos(x) + sin(x)) = (cos(x)/sin(x))*(1/(sin(x))

 

Combine the terms in the numerator on the left and factor out a sin(x) on the denominator on the left. Combine the sin(x) terms on the denominator on the right.

 

((cos(x)+1)/sin(x)) / ((sin(x)(1/cos(x)+1)) = cos(x)/sin^2(x)

 

Combine the fraction in the numerator with the denominator on the left.

 

(cos(x) + 1) / (sin^2(x)*(1/cos(x)+1)) = cos(x)/sin^2(x)

 

Multiply the left side by cos(x)/cos(x)

 

cos(x)*(cos(x) + 1) / (sin^2(x)*(cos(x)+1)) = cos(x)/sin^2(x)
The (cos(x)+1) terms cancel.

 

cos(x)/sin^2(x) = cos(x)/sin^2(x)

 

2. Unless I'm mistaken somewhere, these aren't identities. We can figure that out immediately because sin(x) is defined everywhere but sec(x) and 1/sec(x) have a lot of places where they aren't defined. If you want a full proof, though...

 

We know that sec(x) = 1/cos(x)

 

So

 

1/(1/cos(x)) - 1/cos(x) = -sin(x)

cos(x) - 1/cos(x) = -sin(x)

 

Multiply both sides of the equation by cos(x)

 

(cos^2(x) - 1) = -sin(x)cos(x)

 

Since sin^2(x) + cos^2(x) = 1, cos^2(x) - 1 = -sin^2(x)

 

Substitute:

 

-sin^2(x) = -sin(x)cos(x)

 

-sin(x)*sin(x) = -sin(x)*cos(x)

 

This would mean that sin(x) = cos(x), which isn't true.

 

Another way to see if they're identities is to plot both sides of the equation on the same graph and see if they land on top of each other. In this scenario, they don't.

 

3. Same as always, make everything sine and cosine.

 

tan(x) = sin(x)/cos(x)

cot(x) = cot(x)/sin(x)

 

 

(1 - sin^2(x)/cos^2(x)) / (1 - cos^2(x)/sin^2(x)) = -sin^2(x)/cos^2(x)

 

Combine terms in the numerator and the denominator of the left side by changing 1 into cos^2(x)/cos^2(x) or sin^2(x)/sin^2(x)

 

[(cos^2(x) - sin^2(x))/cos^2(x)] / [(sin^2(x) - cos^2(x))/sin^2(x)] = -sin^2(x)/cos^2(x)

 

Flip the denominator on the left side. You'll be left with two fractions.

 

[sin^2(x) / cos^2(x)]*[(cos^2(x) - sin^2(x)) / [(sin^2(x) - cos^2(x))] = -sin^2(x)/cos^2(x)

 

Factor out negative 1 from the numerator of the second fraction on the left side.

 

-[sin^2(x) / cos^2(x)]*[(sin^2(x) - cos^2(x)) / [(sin^2(x) - cos^2(x))] = -sin^2(x)/cos^2(x)

 

The terms on the left cancel.

 

-[sin^2(x) / cos^2(x)] = -sin^2(x)/cos^2(x)