The value of  $$\frac{\cos 40^\circ}{1 + \sin 40^\circ} + \frac{1 + \sin 40^\circ}{\cos 40^\circ}$$ is:

Correct option is A

Given:

$$\frac{\cos 40^\circ}{1 + \sin 40^\circ} + \frac{1 + \sin 40^\circ}{\cos 40^\circ}$$ is

Formula Used:​

$$\sin^2x+cos^2x=1$$

​$$\frac{1}{cos\theta}=sec\theta$$​​

Solution:

$$= \frac{\cos^2 40^\circ + (1 + \sin 40^\circ)^2}{(1 + \sin 40^\circ) \cos 40^\circ}\\ \ \\= \frac{\cos^2 40^\circ + 1 + 2\sin 40^\circ + \sin^2 40^\circ}{(1 + \sin 40^\circ) \cos 40^\circ}\\ \ \\= \frac{(1 + 1 + 2\sin 40^\circ)}{(1 + \sin 40^\circ) \cos 40^\circ}\\ \ \\$$​

$$= \frac{2(1 + \sin 40^\circ)}{(1 + \sin 40^\circ) \cos 40^\circ}\\ \ \\= \frac{2}{\cos 40^\circ}$$

= 2 sec$$40\degree$$​​

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