If $\frac{\cos θ}{a} = \frac{\sin θ}{b},$ then $\frac{a}{\sec 2 θ} + \frac{b}{cosec 2 θ}$ is equal to

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$a$

$\frac{a}{b}$

$b$

$a + b$

answer is A.

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Detailed Solution

We have,

$\frac{\cos θ}{a} = \frac{\sin θ}{b}$

$\begin{aligned} = \frac{\cos θ}{a} = \frac{\sin θ}{b} = \sqrt{\frac{\cos^{2} θ + \sin^{2} θ}{a^{2} + b^{2}} \frac{-}{b}} \\ = \cos θ = \frac{a}{\sqrt{a^{2} + b^{2}}} and \sin θ = \frac{b}{\sqrt{a^{2} + b^{2}}} \end{aligned}$

$\frac{a}{\sec 2 θ} + \frac{b}{cosec 2 θ} = a \cos 2 θ + b \sin 2 θ$

$\begin{array}{l} = \frac{a}{\sec 2 θ} + \frac{b}{cosec 2 θ} = a (\cos^{2} θ - \sin^{2} θ) + 2 b \sin θ \cos θ \\ = \frac{a}{\sec 20} + \frac{b}{cosec 2 θ} \approx a \frac{(a^{2} - b^{2})}{a^{2} + b^{2}} + \frac{2 a b^{2}}{a^{2} + b^{2}} = a \end{array}$