Can you please help me

Given that sinx=p/q, 00 and q>0

Find cosec2x in terms of p and q.

Can you also add explanation. Thanks :)

Formulas to know:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = 1 - 2sin^2(x)
cos^2(x) + sin^2(x) = 1

The condition 0=0 so the problem is simplified.

Now take all that and use it to rewrite cosec(2x) in terms of sin(x)

cosec(2x) = 1/sin(2x) = 1/(2 sin(x) cos(x)) = 1/(2 sin x sqrt(1 - 2 sin^2 x))

and then substitute sin(x) = p/q

cosec(2x) = 1/(2 p/q sqrt(1 - 2 p^2/q^2))

Simplify by multiplying the numerator and denominator by q^2. One of these factors turns p/q into p and the other one multiplies by q^2 inside the square root

cosec(2x) = q^2 / (2 p sqrt(q^2 - 2 p^2))

cosec(2x)=1/sin(2x)
=1/[2sin(x)cos(x)]

sin(x)=p/q

since x lies in first quadrant, cos(x) is positive.

cos(x)=sqrt(1-sin^2 (x))
=sqrt(1-(p/q)^2)
=sqrt[(q^2 -p^2)/q^2]
=(1/q)sqrt(q^2 -p^2)

Thus,
cosec(2x)=1/[2sin(x)cos(x)]
=1/[2(p/q){(1/q) sqrt(q^2-p^2)}]
=1/[(2p/q^2)sqrt(q^2-p^2)]
=(q^2)/[2p sqrt(q^2-p^2)]

Since cosx = sqrt(1-sin^2x) we have

cosx = sqrt(1-p^2/q^2) = (1/q)sqrt(q^2-p^2)

Now cosec2x = 1/(sin2x) = (1/2)/[sinxcosx]
.....................................= (1/2)/[(p/q)(1/q)sqrt(q^2-p^2)]
.................................... = q^2/[2psqrt(q^2-p2)]

q/(2 p Sqrt[1 - p^2/q^2])