Question

prove that sec +tan = cos/1-sin with steps​

Answer 1

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

Hope this helps:)