(tanx+cotx)over(tanx-cotx)=(1) over sin^2x-cos^2x)
Writing everything in terms of sine and cosine usually works, so ...
LS = (sinx/cosx + cosx/sinx)/(sinx/cosx - cosx/sinx)
= [(sin^2x + cos^2x)/sinxcosx]/[(sin^2x - cos^2x)/(sinxcosx)]
= [(sin^2x + cos^2x)/sinxcosx][(sinxcosx0)/(sin^2x - cos^2x)]
= 1/(sin^2x - cos^2x)
= RS
To solve the given equation (tanx+cotx) / (tanx-cotx) = (1) / (sin^2x-cos^2x), we'll start by simplifying the left-hand side of the equation using the trigonometric identities for tangent and cotangent.
1. Rewrite tanx as sinx/cosx and cotx as cosx/sinx:
(sinx/cosx + cosx/sinx) / (sinx/cosx - cosx/sinx) = 1 / (sin^2x - cos^2x)
2. Find a common denominator for the fractions in the numerator:
[ (sinx)(sinx) + (cosx)(cosx) ] / [ (sinx)(sinx) - (cosx)(cosx) ] = 1 / (sin^2x - cos^2x)
3. Simplify the expressions in the numerator:
[ sin^2x + cos^2x ] / [ sin^2x - cos^2x ] = 1 / (sin^2x - cos^2x)
4. Apply the Pythagorean identity sin^2x + cos^2x = 1:
1 / [ sin^2x - cos^2x ] = 1 / (sin^2x - cos^2x)
5. Since both sides of the equation are the same, the equation is true for all values of x where sin^2x ≠ cos^2x. However, if sin^2x = cos^2x, the equation is undefined.
Therefore, (tanx+cotx) / (tanx-cotx) = 1 / (sin^2x-cos^2x) is true for all values of x where sin^2x ≠ cos^2x.