Question

1/secx-tanx -1/cos = 1/cosx - 1/secx+tanx

Answer 1

Answer:

QED

Step-by-step explanation:

1/(secx-tanx)   -  1/cosx =  1/cosx -  1/(secx+tanx)

LHS = 1/(secx-tanx)   -  1/cosx

= 1 /(1/cosx - sinx/cosx)  - 1/cosx

= cosx/(1-sinx)  - 1/cosx

= cosx(1+sinx)/(1-Sin²x)  - 1/cosx

= cosx(1+sinx)/cos²x  - 1/cosx

= (1+sinx)/cosx - 1/cosx

= (1/cosx)(1 + sinx - 1)

= Sinx/Cosx

= Tanx

RHS = 1/cosx  - 1/(secx + tanx)

= 1/cosx - 1 /(1/cosx + sin/cosx)

= 1/cosx - Cosx/(1+sinx)

=1/cosx - Cosx(1-Sinx)/(1 -Sin²x)

=1/cosx - cosx(1-sinx)/cos²x

=1/cosx - (1-sinx)/cosx

=(1/cosx)(1 -1 +sinx)

= sinx/cosx

= tanx

LHS = RHS

QED

Answer 2

Hence we have proved, the following question.