Question

If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)

Answer 1

Given: Sec θ = x + `1/(4"x"), x ≠ 0,`

Squaring both sides

sec² θ  = `("x" + 1/(4"x"))^2`

We know

tan²θ  = sec²θ - 1

`=>"tan"^2theta = ("x" + 1/("4x"))^2 - 1`

 `=> "tan"^2theta = ("x" + 1/"4x" - 1)("x" + 1/"4x" + 1)`

`=> "tan"^2theta = ("x" - 1/("4x"))^2`

`=> "tan" theta = +- ("x" -1/("4x"))`

When tan θ = x - `1/("4x")`

secθ + tan θ = x + `1/("4x") + "x" - 1/("4x")`

= 2x

When tan θ = - `("x" - 1/("4x")) = 1/("4x") - "x"`

sec θ + tan θ = x + `1/"4x" + 1/"4x" - "x"`

`=1/"2x"`