If sec β + tan β = 2, then what is the value of cot β ?

We are given that ,

sec β + tan β = 2     ----(1)

{ using , sec²β - tan²β = 1  So,  sec β - tan β = \(\frac{1}{sec β + tan β}\) }

sec β - tan β = \(\frac{1}{2}\)     ----(2)

On adding equation 1 and 2 .

2 secβ = 2 + \(\frac{1}{2}\)

2 secβ = \(\frac{5}{2}\)

secβ = \(\frac{5}{4}\)

{ we know, sec A = \(\frac{H}{B}\) }

By using pythagoras theorem,

P² + B² = H²

P² + 4² = 5²

P = 3

Now,

cot β

= \(\frac{B}{P}\)

= \(\frac{4}{3}\)