Question

show that sec x + tan x is equal to root 1 + sin x upon 1 minus sin x​

Answer 1

Step-by-step explanation:

R.H.S=[√{(1+sinx)/(1-sinx)}]×[√{(1+sinx)/(1+sinx)}]. (since rationalizing denominator)

=[√{(1+sinx)(1+sinx)}]/[√{(1-sinx)(1+sinx)}]. (since √a×√b=√ab)

=[√{(1+sinx)^2}]/[√{(1^2)-(sinx^2)}. (since a×a=a^2, (a+b)(a-b)=a^2-b^2 )

=(1+sinx)/[√(1-sinx^2)]. (since √a^2=a)

=(1+sinx)/[√(cosx^2)]. (since 1-sinx^2= cosx^2)

=(1+sinx)/cosx

=(1/cosx)+(sinx/cosx) (since 1/cosx=secx, sinx/cosx=tanx)

=secx+tanx

=L.H.S