if sec theta+tan theta =p ,prove that sin theta = (p square- 1) / (p square + 1)

= Taking R.H.S i.e, (p^{2}-1) / (p^{2}+1)

= Putting value of p in given equation

= [(sec0+tan0)^{2}-1] / [(sec0+tan0)^{2}+1]

= (sec^{2}0+tan^{2}0+2sec0tan0-1) / (sec^{2}0+tan^{2}0+2sec0tan0+1)

= (sec^{2}0-1)+tan^{2}0+2sec0tan0 / (1+tan^{2}0)+sec^{2}0+2sec0tan0

= tan^{2}0+tan^{2}0+2sec0tan0 / sec^{2}0+sec^{2}0+2sec0tan0 [ As sec^{2}0-1=tan^{2}0 1+tan^{2}0=sec^{2}0 ]

= 2tan^{2}0+2sec0tan0 / 2sec^{2}0+2sec0tan0

= 2tan0+(tan0+sec0) / 2sec0+(sec0+tan0) [ Taking 2tan0 2sec0 as common ]

= tan0 / sec0 [ 2 and tan0+sec0 gets cancelled above ]

= sin0/cos0 * /* 1/cos0 [ As tan0=sin0/cos0 and sec0=1/cos0 ]

= sin0/1

= sin0

= L.H.S

Hence Proved...........