If $\sin ^2 \theta-\cos ^2 \theta-3 \sin \theta+2=0,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\frac{1}{\sqrt{\sec \theta-\tan \theta}}$ is:
Answer & explanation
Correct answer: option 1
We are given that :-
sin²θ - cos²θ - 3sinθ + 2 = 0
{ using , sin²θ + cos²θ = 1 }
sin²θ - (1 - sin²θ) - 3sinθ + 2 = 0
2sin²θ - 3sinθ + 1 = 0
2sin²θ - 2sinθ - sinθ + 1 = 0
2sinθ ( sinθ - 1 ) - 1 ( sinθ - 1 ) = 0
( 2sinθ - 1 ).( sinθ - 1 ) = 0
Either ( 2sinθ - 1 ) = 0 or ( sinθ - 1 ) = 0
( sinθ - 1 ) = 0 is not possible because 0º < θ < 90º
So, 2sinθ - 1 = 0
sinθ = \(\frac{1}{2}\)
{ We know, sin30º = \(\frac{1}{2}\) }
So, θ = 30º
Now,
\(\frac{1}{ \sqrt { secθ - tanθ }\}\)
= \(\frac{1}{ \sqrt { sec 30º - tan 30º }\}\)
= \(\frac{1}{ \sqrt { 2/√3 - 1/ √3}\}\)
= \(\frac{1}{ \sqrt { 1/ √3}\}\)'
= $\sqrt[4]{3}$