Fundamental Identities

We have, 

$$f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^6\theta+\sin^6\theta\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^2\theta+\sin^2\theta\right)\left(\cos^4\theta+\cos^2\theta\sin^2\theta+\sin^4\theta\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^4\theta+\cos^2\theta\ \sin^2\theta+\sin^4\theta\right)$$

or, $$f(\theta)=4\cos^4\theta+4\sin^4\theta-2\cos^4\theta+2\cos^2\theta\ \sin^2\theta-2\sin^4\theta$$

or, $$f(\theta)=2\cos^4\theta+2\sin^4\theta+2\cos^2\theta\ \sin^2\theta=2\left(\cos^4\theta+\sin^4\theta+\cos^2\theta\ \sin^2\theta\right)$$

or, $$f(\theta)=2\left(\cos^4\theta+\sin^4\theta+2\cos^2\theta\sin^2\theta-\cos^2\theta\sin^2\theta\right)=2\left(1-\cos^2\theta\sin^2\theta\right)$$

or, $$f(\theta)=2\left(1-\cos^2\theta\ \sin^2\theta\right)=2-2\cos^2\theta\ \sin^2\theta=2-\dfrac{\sin^22\theta}{2}\ $$

This expression has maximum and minimum values as $$\alpha$$ and $$\beta$$ respectively.

The maximum and minimum value for $$\sin^22\theta$$ is 1 and 0 respectively for $$\theta\in R\ $$. 

Hence, 

$$f(\theta)_{\min}=2-\dfrac{1}{2}=\dfrac{3}{2}\ =\beta\ $$

$$f(\theta)_{\max}=2-\dfrac{0}{2}=2\ =\alpha\ $$

So, $$\alpha+2\beta\ =2+2\times\dfrac{3}{2}=2+3=5$$

Let $$\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$$ and $$\beta = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$$. The expression becomes $$\tan\left(2(\alpha - \beta)\right)$$.

First, find $$\tan\alpha$$ and $$\tan\beta$$.

For $$\alpha$$:
Given $$\sin\alpha = \frac{2}{\sqrt{13}}$$, use $$\sin^2\alpha + \cos^2\alpha = 1$$:
$$\cos^2\alpha = 1 - \left(\frac{2}{\sqrt{13}}\right)^2 = 1 - \frac{4}{13} = \frac{9}{13}$$.
Since $$\alpha$$ is in the first quadrant (principal value of $$\sin^{-1}$$), $$\cos\alpha = \frac{3}{\sqrt{13}}$$.
Thus, $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{2/\sqrt{13}}{3/\sqrt{13}} = \frac{2}{3}$$.

For $$\beta$$:
Given $$\cos\beta = \frac{3}{\sqrt{10}}$$, use $$\sin^2\beta + \cos^2\beta = 1$$:
$$\sin^2\beta = 1 - \left(\frac{3}{\sqrt{10}}\right)^2 = 1 - \frac{9}{10} = \frac{1}{10}$$.
Since $$\beta$$ is in the first quadrant (principal value of $$\cos^{-1}$$), $$\sin\beta = \frac{1}{\sqrt{10}}$$.
Thus, $$\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$$.

Now, find $$\tan(\alpha - \beta)$$ using the tangent subtraction formula:
$$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} = \frac{\frac{2}{3} - \frac{1}{3}}{1 + \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)} = \frac{\frac{1}{3}}{1 + \frac{2}{9}} = \frac{\frac{1}{3}}{\frac{11}{9}} = \frac{1}{3} \times \frac{9}{11} = \frac{3}{11}$$.

Let $$\theta = \alpha - \beta$$, so $$\tan\theta = \frac{3}{11}$$. Now find $$\tan(2\theta)$$ using the double-angle formula:
$$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2 \times \frac{3}{11}}{1 - \left(\frac{3}{11}\right)^2} = \frac{\frac{6}{11}}{1 - \frac{9}{121}} = \frac{\frac{6}{11}}{\frac{112}{121}} = \frac{6}{11} \times \frac{121}{112} = \frac{6 \times 121}{11 \times 112}$$.

Simplify:
$$\frac{6 \times 121}{11 \times 112} = \frac{6 \times 11 \times 11}{11 \times 112} = \frac{6 \times 11}{112} = \frac{66}{112}$$.
Reduce by dividing numerator and denominator by 2:
$$\frac{66 \div 2}{112 \div 2} = \frac{33}{56}$$.

Since $$\alpha$$ and $$\beta$$ are both acute and $$\tan\alpha = \frac{2}{3} > \tan\beta = \frac{1}{3}$$, it follows that $$\alpha > \beta$$, so $$\theta = \alpha - \beta > 0$$. Thus, $$2\theta$$ is in the first quadrant where tangent is positive.

The value is $$\frac{33}{56}$$, which corresponds to option A.

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 $$2$$.