Question

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

Answer 1

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)`