Prove that the following Trigonometric Identity is valid:
1/(sin(y)-1)-1/(sin(y)+1)=-2sec^2 (y)
Please any help or clues on how to do this is needed. I have a sheet that tells me valid identities but nowhere on it has info on -2sec^2 (y) I am sooo clueless here :(
LS = 1/(siny - 1) - 1/(siny + 1)
using a common denominator:
= (siny + 1 - (siny - 1)/( sin^2 y - 1)
= 2/(-cos^2 y)
= - 2 sec^2 y
= RS
THANK YOU REINY!!! however, what does RS mean?
LS ---> left side
RS ---> ???
ooohkay, thank you!! i needed this so bad :D
To prove the given trigonometric identity, we'll start with the left-hand side (LHS) and simplify it until it matches the right-hand side (RHS). Here's a step-by-step explanation:
1. Start with the LHS: 1/(sin(y) - 1) - 1/(sin(y) + 1).
2. The first step is to find a common denominator for the two fractions. To do this, multiply the first fraction by (sin(y) + 1) and the second fraction by (sin(y) - 1). This results in:
[(sin(y) + 1) - (sin(y) - 1)] / [(sin(y) - 1)(sin(y) + 1)].
Simplifying further:
= [sin(y) + 1 - sin(y) + 1] / [(sin^2(y) - 1)].
= [2] / [(sin^2(y) - 1)].
3. Now, we'll simplify the denominator by using the Pythagorean identity: sin^2(y) = 1 - cos^2(y). Replacing sin^2(y) in the denominator:
= 2 / [(1 - cos^2(y)) - 1].
= 2 / [-cos^2(y)].
= -2 / [cos^2(y)].
4. Finally, we'll use the reciprocal identity: sec^2(y) = 1 / cos^2(y). Replacing -2 / [cos^2(y)] with the RHS:
= -2 * 1 / [cos^2(y)].
= -2sec^2(y).
Therefore, we have successfully shown that the LHS is equal to the RHS, proving the given trigonometric identity: 1/(sin(y) - 1) - 1/(sin(y) + 1) = -2sec^2(y).