Practicing Success
If 4 - 2sin2θ - 5cosθ = 0, 0° < θ < 90°, then find sinθ+ tanθ. |
\(\frac{3\sqrt {2}}{2}\) \(\frac{3\sqrt {3}}{2}\) 3\(\sqrt {2}\) 2\(\sqrt {3}\) |
\(\frac{3\sqrt {3}}{2}\) |
4 - 2sin2θ - 5cosθ = 0 Put θ = 60° ⇒ 4 - 2 × (\(\frac{\sqrt{3}}{2}\))2 - \(\frac{5}{2}\) ⇒ 4 - \(\frac{6}{4}\) - \(\frac{5}{2}\) = 0 satisfied So, ⇒ tanθ+ sinθ = \(\sqrt {3}\) + \(\frac{\sqrt {3}}{2}\) = \(\frac{3\sqrt {3}}{2}\) |