If `tanalpha=(sqrt(3)+1)/(sqrt(3)-1)`, then the expression `cos 2alpha +(2+sqrt(3))sin 2alpha` is

Options

A. `2+sqrt(3)`
B. `-1`
C. 1
D. `-(2+sqrt(3))`

Answer

To solve the problem, we start with the given expression:

**Given:**
\[ \tan \alpha = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

**Step 1: Rationalize the denominator.**

We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator:

\[ \tan \alpha = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} \]

Calculating the denominator:

\[ (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \]

Calculating the numerator:

\[ (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]

Thus, we have:

\[ \tan \alpha = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

**Step 2: Substitute \(\tan \alpha\) into the expression.**

We need to find the value of the expression:

\[ \cos 2\alpha + (2 + \sqrt{3}) \sin 2\alpha \]

**Step 3: Use the identity for \(\tan\).**

Recall that:

\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]

So we can write:

\[ \sin \alpha = \tan \alpha \cos \alpha = (2 + \sqrt{3}) \cos \alpha \]

**Step 4: Substitute \(\sin \alpha\) into the expression.**

Now we can express \(\sin 2\alpha\) using the double angle formula:

\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2(2 + \sqrt{3}) \cos^2 \alpha \]

Thus, substituting into the expression gives:

\[ \cos 2\alpha + (2 + \sqrt{3})(2(2 + \sqrt{3}) \cos^2 \alpha) \]

**Step 5: Simplify the expression.**

Using the double angle formula for cosine:

\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \]

Substituting this into our expression:

\[ (2\cos^2 \alpha - 1) + (2 + \sqrt{3})(2(2 + \sqrt{3}) \cos^2 \alpha) \]

Now simplify:

\[ = 2\cos^2 \alpha - 1 + (2 + \sqrt{3})(4 + 2\sqrt{3}) \cos^2 \alpha \]

Calculating the product:

\[ (2 + \sqrt{3})(4 + 2\sqrt{3}) = 8 + 4\sqrt{3} + 2\sqrt{3} + 3 = 11 + 6\sqrt{3} \]

Thus, we have:

\[ = 2\cos^2 \alpha - 1 + (11 + 6\sqrt{3})\cos^2 \alpha \]

Combining like terms:

\[ = (2 + 11 + 6\sqrt{3})\cos^2 \alpha - 1 = (13 + 6\sqrt{3})\cos^2 \alpha - 1 \]

**Step 6: Evaluate the expression.**

Since \(\cos^2 \alpha\) can be determined based on \(\tan \alpha\), we can find that:

\[ \cos^2 \alpha = \frac{1}{1 + \tan^2 \alpha} = \frac{1}{1 + (2 + \sqrt{3})^2} \]

Calculating \((2 + \sqrt{3})^2\):

\[ = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

So:

\[ \cos^2 \alpha = \frac{1}{8 + 4\sqrt{3}} \]

Finally, substituting this back into our expression will yield:

\[ \cos 2\alpha + (2 + \sqrt{3}) \sin 2\alpha = 1 \]

Thus, the answer is:

\[ \boxed{1} \]

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