If $\sqrt{2} sec^2 θ - 4 sec θ +2\sqrt{2} = 0, $ then what is the value $sin^2 θ + tan^2 θ $ ?

3/2

We are given that :-

√2 sec²θ - 4secθ + 2√2 = 0 

√2 sec²θ - 2secθ -  2secθ + 2√2 = 0 

√2 secθ ( secθ - √2 ) - 2 ( secθ - √2 ) = 0

(√2 secθ - 2 ). ( secθ - √2 ) = 0

Either (√2 secθ - 2 ) = 0 or ( secθ - √2 ) = 0

(√2 secθ - 2 ) = 0 is not possible

So, ( secθ - √2 ) = 0 

secθ = √2

{ we know, sec 45º = √2 }

So, θ = 45º

Now,

sin²θ + tan²θ

= sin²45º + tan²45º

= \(\frac{1}{2}\) + 1

= \(\frac{3}{2}\)