(Sec inverse p) - (cosec inverse q) =0 then p^-2-q^-2=?
Let's simplify the given equation first:
(sec^(-1)(p)) - (cosec^(-1)(q)) = 0
This can be rewritten as:
arccos(1/p) - arcsin(1/q) = 0
Since arccos(1/p) and arcsin(1/q) are angles whose cosine and sine values are 1/p and 1/q respectively, we have:
cos(arccos(1/p)) = 1/p and sin(arcsin(1/q)) = 1/q
cos(arccos(1/p)) = 1/p simplifies to:
1/p = 1/p
sin(arcsin(1/q)) = 1/q simplifies to:
1/q = 1/q
Therefore, we have 1/p = 1/p and 1/q = 1/q which are always true.
Now let's solve for p^(-2) - q^(-2):
p^(-2) - q^(-2) = 1/p^2 - 1/q^2 = (q^2 - p^2)/(p^2*q^2) = (q + p)(q - p)/(p^2*q^2)
Given that sec^(-1)(p) - cosec^(-1)(q) = 0, this implies that q + p = 0 or q = -p. Substituting q = -p into p^(-2) - q^(-2) = (q + p)(q - p)/(p^2*q^2), we get:
p^(-2) - (-p)^(-2) = p^(-2) - p^(-2) = 0
Therefore, p^(-2) - q^(-2) = 0.