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We'll raise to square the 1st constraint sin x+cos x=p.
(sin x+cos x )^2= p^2
We'll expand the square:
(sin x)^2 + (cos x)^2 + 2sin x*cos x = p^2
We'll apply the Pythagorean identity for (sin x)^2 + (cos x)^2 = 1.
1 + 2sin x*cos x = p^2 (1)
We'll use the fact that tan x = sin x/cos x and cot x = cos x/sin x
We'll re-write the 2nd constraint:
sin x/cos x + cos x/sin x = q
[(sin x)^2 + (cos x)^2]/sin x*cos x = q
1/sin x*cos x = q
sin x*cos x = 1/q (2)
We'll substitute (2) in (1):
1 + 2/q= p^2
We'll multiply by q both sides:
q + 2 = q*p^2
We'll substract q both sides:
q*p^2 - q = 2
q(p^2 - 1) = 2
The relation between p and q, ifsin x+cos x=p and tan x+cot x=q, is: q(p^2 - 1) = 2.
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