If $sin^2 \theta - cos^2\theta - 3 sin \theta + 2 = 0, 0^o < \theta < 90^o$, then what is the value of $1 + sec\theta +tan\theta $?
Answer & explanation
Correct answer: option 3
sin2θ - cos2θ - 3 sinθ + 2 = 0
sin2θ - ( 1 - sin2θ) - 3 sinθ + 2 = 0
2sin2θ - 3 sinθ + 1 = 0
on solving ,
sinθ = \(\frac{1}{2}\)
so , θ = 30º
Now , 1 + secθ + tanθ
= 1 + sec30º + tan30º
= 1 + \(\frac{2}{√3}\) + \(\frac{1}{√3}\)
= \(\frac{√3 + 2 + 1}{√3}\)
= 1 + √3