The value of $$\frac{\sin^2 16^\circ}{\cos^2 74^\circ} + \frac{\tan^2 14^\circ}{\cot^2 76^\circ}$$​= is:

Correct option is A

Given : 

$$\frac{\sin^2 16^\circ}{\cos^2 74^\circ} + \frac{\tan^2 14^\circ}{\cot^2 76^\circ}$$ ,  

Formula Used: 

$$\cos(90-A) = \sin A \\ \tan(90-A) = \cot A$$​

Solution : 

Simplify each term :

$$\frac{\sin^2 16^\circ}{\cos^2 74^\circ}$$​     

Using the complementary angle identify, 

​$$cos^2 74^\circ$$​  = $$\sin^2 16^\circ$$ 

$$\frac{\sin^2 16^\circ}{\cos^2 74^\circ} = \frac{\sin^2 16^\circ}{\sin^2 16^\circ}$$  = 1.

  $$\frac{\tan^2 14^\circ}{\cot^2 76^\circ}$$ 

$$\frac{\tan^2 14^\circ}{\cot^2 76^\circ} = \frac{\tan^2 14^\circ}{\tan^2 14^\circ}= 1$$  

​$$\cot^2 76^\circ = \tan^2 14^\circ$$​​

Hence, 

​$$\frac{\sin^2 16^\circ}{\cos^2 74^\circ} + \frac{\tan^2 14^\circ}{\cot^2 76^\circ} = 1+ 1 = 2$$   

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