If cos A + sec A = $$\frac{5}{2}$$​ and A is an acute angle, then sin A + cosecA is equal to:

Correct option is C

Given:

​$$\cos A + \sec A = \frac{5}{2}$$​​

A is an acute angle ( $$0 < A < 90^\circ )$$​
Formula Used:

$$\sec A = \frac{1}{\cos A}$$​​

$$\cosec A = \frac{1}{\sin A}$$​​

​$$\sin^2 A + \cos^2 A = 1$$​​
Solution:

Let cos A = x 
Given:
​$$x + \frac{1}{x} = \frac{5}{2}$$​

$$2x^2 + 2 = 5x$$

​​$$2x^2 - 5x + 2 = 0$$​

​$$x = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$$​

​$$x = \frac{5 + 3}{4} = 2 \\ \ \\x = \frac{5 - 3}{4} = \frac{1}{2} \\ \ \\\cos A = \frac{1}{2}$$

Since A is an acute angle,$$\cos A = \frac{1}{2}$$​(since cos A = 2 is not possible for an acute angle).​​

$$\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{2}\right)^2 = \frac{3}{4}$$​​

Since A is acute, sin A is positive:

$$\sin A = \frac{\sqrt{3}}{2} \\ \ \\\cosec A = \frac{1}{\sin A} = \frac{2}{\sqrt{3}} \\ \ \\\sin A + \cosec A = \frac{\sqrt{3}}{2} + \frac{2\sqrt{3}}{3} = \frac{3\sqrt{3} + 4\sqrt{3}}{6} = \frac{7\sqrt{3}}{6} = \frac{7}{2\sqrt3}$$

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