What is the identity in p and q, if sin x+cos x=p and tan x+cot x=q?

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.