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 $?

$ 1+\sqrt{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