If secA-tanA=a,show that ; secA=1/2[a+1/a]& tanA=1/2[1/a-a]?

1 Answer
Feb 17, 2018

Please look explanation section.

Explanation:

secA-tanA=a

(secA-tanA)(secA+tanA)=a(secA+tanA)

a*(secA+tanA)=(secA)^2-(tanA)^2

a*(secA+tanA)=1

secA+tanA=1/a

After solving equations of secA-tanA=a and secA+tanA=1/a are together,

secA=1/2*(a+1/a) and tanA=1/2*(1/a-a)