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If $$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }},$$ 0° < θ < 90°, then the value of secθ + tanθ is:
Answer & Solution
Correct Answer:
Option
A
$$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }}$$
$$\eqalign{ & \sec \theta + \tan \theta \cr & = \frac{{\sqrt {{a^2} + {b^2}} }}{b} + \frac{a}{b} \cr & = \frac{{\sqrt {{a^2} + {b^2}} + a}}{b} \cr} $$
$$\eqalign{ & \sec \theta + \tan \theta \cr & = \frac{{\sqrt {{a^2} + {b^2}} }}{b} + \frac{a}{b} \cr & = \frac{{\sqrt {{a^2} + {b^2}} + a}}{b} \cr} $$
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