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 :
Answer & explanation
Correct answer: option 2
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\(\frac{2}{3}\)