If 3 cos ^2 θ-4 sin θ+1=0,0^°<θ<90^°, then tan θ+sec θ= ?

We are given that :-

3 cos²θ - 4sinθ + 1 = 0

{ we know, sin²θ  + cos²θ  = 1 }

3 ( 1 - sin²θ ) - 4sinθ + 1 = 0

3sin²θ + 4sinθ - 4 = 0  = 0

3sin²θ + 6sinθ - 2sinθ - 4 = 0  = 0

3sinθ ( sinθ + 2 ) - 2 ( sinθ + 2 ) = 0

( 3sinθ - 2 ).( sinθ + 2 ) = 0

Either ( 3sinθ - 2 ) = 0 or ( sinθ + 2 ) = 0 

If ( sinθ + 2 ) = 0

Then sinθ = -2 is not possible.

So,  3sinθ - 2  = 0

sinθ = \(\frac{2}{3}\)

{ we know, sinA = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

2² + B² = 3²

B = √5

Now,

tanθ + secθ

= \(\frac{P}{B}\) + \(\frac{H}{B}\)

= \(\frac{2}{√5}\) + \(\frac{3}{√5}\)

= \(\frac{5}{√5}\)

= √5