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.