Question

Prove that sec + tan = cos/1-sin​

Answer 1

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Answer:

LHS=secA+tanA RHS= cos/(1-sinA)

Step-by-step explanation:

Taking, LHS=

as, {secA=1/cosA} and {tanA=sinA/cosA}

=(1/cosA)+(sinA/cosA)

=(1+sinA)/cosA

on Rationalising denominator;

=[(1+sinA)/cosA]×[cosA/cosA]

=[(1+sinA)cosA]÷cos²A

as, sin²A+cos²A=1

=[(1+sinA)cosA]÷(1-sin²A)

using identity, a²–b²=(a+b)(a-b)

=[(1+sinA)cosA]÷[(1+sinA)(1-sinA)]

=cos/1-sinA

Hence, proved

Answer 2

Answer:

hey buddy

taking R.H.S.

We have, cosx.

1−sinx ,

= cosx. × 1,

1−sinx

= cosx. × 1 + sinx

1−sinx. 1 + sinx

= cosx( 1 + sinx )

1−sin^2x

= 1 + sinx

cosx

= 1. +. sinx

cosx. cosx

= secx + tanx, as desired!

hope it helped uhh