The value of $$\frac{1}{(1+\cot^2 x)^2} + \frac{\tan^2 x}{(1+\tan^2 x)^2} + \frac{1}{1+\tan^2 x}$$​ is:

Correct option is D

Given:

​$$\frac{1}{(1+\cot^2 x)^2} + \frac{\tan^2 x}{(1+\tan^2 x)^2} + \frac{1}{1+\tan^2 x}$$

Concept Used:

trigonometric identities

(1). 1 + tan²​$$\theta$$ = sec²​$$\theta$$​​

(2). 1 + cot²​$$\theta$$ = cosec²​$$\theta$$ 

(3) $$\tan x = \frac{\sin x}{\cos x}$$, $$\cos x = \frac{1}{\sec x}$$​​

Solution: 

Applying the identities;

= $$\frac{1}{cosec^4x} + \frac{\tan^2 x}{sec^4x} + \frac{1}{sec^2x}$$

= $${\sin}^4x+{\cos}^4x\ \times\frac{{\sin}^2x}{{\cos}^2x}+{\cos}^2x$$

= sin²x (sin²x + cos^2​x) + cos²x    (sin²x + cos²x)   .......... (first identity)

= 1