​Solve the following.

$$\frac{\tan A}{1 + \sec A} + \frac{1 + \sec A}{\tan A} = ?$$​​

Correct option is C

Given:

​$$\frac{\tan A}{1 + \sec A} + \frac{1 + \sec A}{\tan A}$$ 

Formula Used:

​$$\sec A = \frac{1}{\cos A}$$ 

​$$\tan A = \frac{\sin A}{\cos A}$$ 

Pythagorean identity $$sin^2 A + \cos^2 A = 1$$​​

Solution:

Substitute identity into the expression:

​$$\frac{\frac{\sin A}{\cos A}}{1 + \frac{1}{\cos A}} + \frac{1 + \frac{1}{\cos A}}{\frac{\sin A}{\cos A}}$$ 

= $$\frac{\frac{\sin A}{\cos A}}{\frac{\cos A + 1}{\cos A}} + \frac{\frac{\cos A + 1}{\cos A}}{\frac{\sin A}{\cos A}}$$ 

= $$\frac{\sin A}{\cos A + 1} + \frac{\cos A + 1}{\sin A}$$ 

= $$\frac{\sin^2 A + (\cos A + 1)^2}{(\cos A + 1) \sin A}$$  

= $$\frac{2(1 + \cos A)}{(1 + \cos A) \sin A}$$ 

= $$\frac{2}{\sin A}$$

= 2 cosec A 

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