Tanx+secx-1÷tanx-secx+1 prove thatdy÷dx= sec (secx+tanx)

(tanx+secx−1) / (tanx−secx+1)

Multiplying numerator and denominator by tanx+secx+1

= {(tanx+secx−1) / (tanx−secx+1)} × {(tanx+secx+1) / (tanx+secx+1)}

= {tan² x+tanxsecx+tanx+tanxsecx+sec² x+secx−tanx−secx−1}/

{tan² x+tanxsecx+tanx−tanxsecx−sec² x−secx+tanx+secx+1}

= (tan² x+2tanxsecx+sec² x−1) / (tan² x+2tanx−sec² x+1)

As sec2x=tan² x+1, above is equal to

= (tan² x+2tanxsecx+tan² x+1−1) / (tan² x+2tanx−tan² x−1+1)

= (2tan² x+2tanxsecx) / 2tanx

= 2tanx(tanx+secx) / 2tanx

= tanx+secx

Now differentiating above wrt x

d/dx(tanx) + d/dx(secx)

=> sec (secx+tanx)