Differentiate 1/tanx
I got -sec^2 / tan^2x, which is correct, but the answers said I could also simplify this to -cosec^2. How would you do that?
Thanks =)
-sec^2/tan^2 = -sec^2 * cos^2/sin^2 = -1/sin^2 = -csc^2
also, just as
tan' = sec^2
cot' = -csc^2 as you proved
To simplify -sec^2/tan^2x to -cosec^2x, you need to use trigonometric identities.
First, let's rewrite -sec^2x in terms of sine and cosine. Recall that secx is equal to 1/cosx:
-sec^2x = - (1/cos^2x)
Now, let's rewrite tan^2x in terms of sine and cosine. Recall that tanx is equal to sinx/cosx:
tan^2x = (sin^2x)/(cos^2x)
Now, substitute these values back into the original expression:
-sec^2x/tan^2x = - (1/cos^2x) / ((sin^2x)/(cos^2x))
When you divide fractions, you multiply by the reciprocal of the second fraction. So, we can rewrite the expression as:
- (1/cos^2x) * (cos^2x/(sin^2x))
Now, simplify by canceling out the common terms:
= - (1/sin^2x)
The expression - (1/sin^2x) is equivalent to -cosec^2x. Therefore, the simplified form of -sec^2x/tan^2x is -cosec^2x.
I hope this helps! Let me know if you have any further questions.