trig identites

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britttt

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Mar 11, 2013
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I need some help with my math homework....we're verifying trig identities. Here's the problem...

tan^2x(cot^2x-cos^2x)
 
What you posted is not an identity; it is a trigonometric expression.

Please post the exact instructions, for this expression.

Cheers ~ Mark :cool:
 
Rule #1 Try somethnig - Anything!

What have you tried?

Rule #2 When in doubt, change everythin gto sine and cosine. It is likely you are moer familiar with it, after that.

Rule #3 Seriously, just jump in and try something.
 
the expression tan^2x(cot^2x-cos^2x) is equivalent to:

a. cot^2x
b. tan^2x
c. sin^2x
d. cos^sx
 
Hmmm, with some scratch-work, I note that the given expression simplifies to the square of an elementary trig function.

Instead of verifying an identity, are you perhaps supposed to simplify the given expression?

If so, start by rewriting in terms of sines and cosines only.
 
britttt said:
the expression tan^2x(cot^2x-cos^2x) is equivalent to:

a. cot^2x
b. tan^2x
c. sin^2x
d. cos^sx
Click to expand...

We cross-posted.

Yes, one of those multiple-choices is equivalent to the given expression.

Do you know how to rewrite tan(x) and cot(x) in terms of sines and cosines?
 
Very good. Now replace the squares of tangent and cotangent in the exercise with squares of sin/cos and cos/sin, appropriately. Then, look for cancellations between numerators and denominators, when doing the given multiplication.
 
In lght of Mrak's remarks,...

Rule #0 Make sure it's an identity. Maybe you can simplify the expreession. :)
 
sin (x)/ cos (x) (cos (x)/ sin (x) -cos^2x)....do both of the sin and cos cancel?
 
Here's a quick review of the Distributive Property:

\(\displaystyle \frac{S^2}{C^2}(\frac{C^2}{S^2} - \frac{C^2}{1}) = \frac{S^2}{C^2} \cdot \frac{C^2}{S^2} - \frac{S^2}{C^2} \cdot \frac{C^2}{1}\)

Look for cancellations, on the right-hand side. :cool:

​
 
I would, but the library is closing.

You may begin as many new threads (one for each new exercise) as you like!

Others will guide you. Cheers ~ Mark :cool:
 
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