If $3 cos^2 \theta - 4 sin \theta + 1 = 0, 0° < \theta < 90°$ , then the value of $3 cos^2 \theta + 5tan^2 \theta $ will be :

$5\frac{2}{3}$

3cos²θ - 4sinθ + 1 = 0

3( 1 - sin²θ ) - 4sinθ + 1 = 0

3 - 3sin²θ  - 4sinθ + 1 = 0

3sin²θ  + 4sinθ - 4 = 0

3sin²θ  + 6sinθ  - 2sinθ - 4 = 0

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

( 3sinθ - 2 ). ( sinθ + 2 ) = 0

Either 3sinθ - 2 = 0 Or sinθ + 2 = 0

sinθ + 2 = 0 is not possible as θ is b/w 0º and 90º

So, 3sinθ - 2 = 0

sinθ = \(\frac{2}{3}\)

{ sinθ = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

2² + B² = 3²

B = √5

Now,

3cos²θ + 5tan²θ

= 3 × \(\frac{5}{9}\) + 5 × \(\frac{4}{5}\)

=  \(\frac{15}{9}\) + 4

= \(\frac{51}{9}\)

= 5\(\frac{2}{3}\)