If sec θ - tan θ = 3, then cos θ is equal to :

$\frac{3}{5}$

Using :-

sec²x - tan²x = 1

So, secx - tanx = \(\frac{1 }{secx +tanx}\)

Given :-

sec θ - tan θ = 3     ---(1)

So, secθ + tanθ = \(\frac{1 }{secθ -tanθ}\)

= \(\frac{1 }{3}\)    ----(2)

Adding 1 and 2.

2sec θ = 3 + \(\frac{1 }{3}\) 

sec θ = \(\frac{5 }{3}\) 

{ sec θ = \(\frac{H }{B}\)  }

using  pythagoras theorem,

P² + B² = H²

P² + 3² = 5²

P = 4

Now,

cos θ

= \(\frac{B}{H}\)  

= \(\frac{3 }{5}\)