sorry, another I can't figure out
Show that (1-cot^2x)/(tan^2x-1)=cot^2x
I started by factoring both as difference of squares. Would I be better served by writing in terms of sine and cosine? Such as:
[1-(cos^2x/sin^2x)]/[(sin^2x/cos^2x)-1]=(cos^2x/sin^2x)
I don't think your equation is an identity,
I tried several angles and the Left Side is not equal to the Right Side.
Check your typing.
In general, I try to prove these type of identities by changing everything to sines and cosines, unless I can recognize one of the common trig relationships.
let me try to type it out again, my apologies.
Original problem is--Prove the following:
1- cot^2 x
----------- = cot^2 x
tan^2 x - 1
That is how the problem actually looks.
To prove the given identity: (1 - cot^2x) / (tan^2x - 1) = cot^2x
We can start by expressing the terms in the given expression in terms of sin(x) and cos(x).
Step 1: Rewrite cot^2x, tan^2x, and 1 in terms of sin(x) and cos(x):
cot^2x = (cos^2x / sin^2x)
tan^2x = (sin^2x / cos^2x)
1 = (sin^2x / sin^2x) (cos^2x / cos^2x) = (sin^2x + cos^2x) / (sin^2x + cos^2x) = 1
Now, let's substitute these expressions back into the original equation:
(1 - cot^2x) / (tan^2x - 1) = [(1 - (cos^2x / sin^2x)) / ((sin^2x / cos^2x) - 1)]
Step 2: Simplify the left side of the equation:
(1 - cot^2x) = 1 - (cos^2x / sin^2x)
= (sin^2x - cos^2x) / sin^2x
Step 3: Simplify the right side of the equation:
(tan^2x - 1) = (sin^2x / cos^2x) - 1
= (sin^2x - cos^2x) / cos^2x
Now, we can substitute these simplified expressions back into the equation:
[(1 - (cos^2x / sin^2x)) / ((sin^2x / cos^2x) - 1)]
= [(sin^2x - cos^2x) / sin^2x] / [(sin^2x - cos^2x) / cos^2x]
= (sin^2x - cos^2x) / sin^2x * cos^2x / (sin^2x - cos^2x)
= (sin^2x - cos^2x) * cos^2x / (sin^2x - cos^2x)
Step 4: Cancel out the (sin^2x - cos^2x) terms:
= cos^2x
Therefore, we have shown that (1 - cot^2x) / (tan^2x - 1) is equal to cot^2x.