Practicing Success
If $3 \cos ^2 \theta-4 \sin \theta+1=0,0^{\circ}<\theta<90^{\circ}$, then $\tan \theta+\sec \theta=$ ? |
$2 \sqrt{5}$ $2 \sqrt{3}$ $3 \sqrt{3}$ $\sqrt{5}$ |
$\sqrt{5}$ |
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 |