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

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

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago

(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)

 

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