If $$\sin x + \sin^2 x = 1$$, $$x \in (0, \tfrac{\pi}{2})$$ then $$(\cos^{12}x+\tan^{12}x)+3(\cos^{10}x+\tan^{10}x+\cos^{8}x+\tan^{8} x)+(\cos^{6}x+\tan^{6}x)$$
is equal to :

Step 1: Simplify the given trigonometric condition

Given the equation:

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

We know the fundamental trigonometric identity:

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

By comparing these two equations, we can deduce:

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

Step 2: Establish a relationship between tangent and cosine

Using the fundamental definition of the tangent function:

$$\tan x = \frac{\sin x}{\cos x}$$

Substitute $$\sin x = \cos^2 x$$ into the numerator:

$$\tan x = \frac{\cos^2 x}{\cos x}$$
$$\tan x = \cos x$$

This implies that for any integer power $$k$$:

$$\tan^k x = \cos^k x$$

Step 3: Substitute and simplify the main expression

The expression we need to evaluate is:

$$(\cos^{12} x + \tan^{12} x) + 3(\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$$

Replace all $$\tan^k x$$ terms with $$\cos^k x$$:

$$(\cos^{12} x + \cos^{12} x) + 3(\cos^{10} x + \cos^{10} x + \cos^8 x + \cos^8 x) + (\cos^6 x + \cos^6 x)$$

Combine the like terms:

$$2\cos^{12} x + 3(2\cos^{10} x + 2\cos^8 x) + 2\cos^6 x$$
$$2\cos^{12} x + 6\cos^{10} x + 6\cos^8 x + 2\cos^6 x$$

Factor out the common multiplier $$2$$:

$$2(\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x)$$

Step 4: Condense the algebraic expression

Notice that the terms inside the parentheses form a perfect cube algebraic expansion, which resembles the standard identity $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$.

Let $$a = \cos^4 x$$ and $$b = \cos^2 x$$. We can verify this setup:

• $$a^3 = (\cos^4 x)^3 = \cos^{12} x$$
• $$3a^2b = 3(\cos^4 x)^2(\cos^2 x) = 3(\cos^8 x)(\cos^2 x) = 3\cos^{10} x$$
• $$3ab^2 = 3(\cos^4 x)(\cos^2 x)^2 = 3(\cos^4 x)(\cos^4 x) = 3\cos^8 x$$
• $$b^3 = (\cos^2 x)^3 = \cos^6 x$$

Since all terms match perfectly, the expression condenses to:

$$2(\cos^4 x + \cos^2 x)^3$$

Step 5: Final substitution and evaluation

From Step 1, we established $$\cos^2 x = \sin x$$. Squaring both sides yields $$\cos^4 x = \sin^2 x$$.

Substitute these values back into our condensed expression:

$$2(\sin^2 x + \sin x)^3$$

Since the original problem states $$\sin x + \sin^2 x = 1$$, we can substitute $$1$$ into the parenthesis:

$$2(1)^3 = 2(1) = 2$$

Final Answer

The value of the given expression is $$2$$.