6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1 ... ?

I will use the letter x instead of theta.

Then the problem is, given sec (x) + tan (x) = P, show that

sin (x) = [P^2 - 1] / [P^2 + 1]

I am going to take a non regular path.

First, develop a little the left side of the first equation:

sec (x) + tan (x) = 1 / cos (x) + sin (x) / cos (x) = [1 + sin (x) ] / cos (x)

and that is equal to P.

Second, develop the rigth side of the second equation:

[p^2 - 1] / [p^2 + 1] =

= [ { [1 + sin (x) ] / cos (x) }^2 - 1] / [ { [1 + sin (x) ] / cos (x) }^2 + 1 ] =

= { [1 + sin (x) ]^2 - [cos (x) ]^2 } / { [1 + sin (x) ]^2 + [cos (x) ]^2 } =

= {1 + 2sin (x) + [sin (x) ^2] - [cos (x) ^2] } / {1 + 2sin (x) + [sin (x) ^2] + [cos (x) ^2] }

= {2sin (x) + [sin (x) ]^2 + [sin (x) ]^2 } / { 1 + 2 sin (x) + 1} =

= {2sin (x) + 2 [sin (x) ]^2 } / {2 + 2sin (x) } = {2sin (x) (1 + sin (x) } / {2 (1+sin (x) } =

= sin (x)

Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin (x), the second equation.