If $\sqrt{2} sec^2 θ - 4 sec θ +2\sqrt{2} = 0, $ then what is the value $sin^2 θ + tan^2 θ $ ?
Answer & explanation
Correct answer: option 4
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
= \(\frac{3}{2}\)