Find the exact value of the following expression.

Find the exact value of the following expression. \tan\left(\frac{\pi}{6} + \frac{5\pi}{4}\right) Rewrite the expression using a sum or difference formula. Choose the correct answer below. \frac{\tan\frac{5\pi}{4} – \tan\frac{\pi}{6}}{1 – \tan\frac{\pi}{6}\tan\frac{5\pi}{4}} A. \frac{\tan\frac{\pi}{6} + \tan\frac{5\pi}{4}}{1 – \tan\frac{\pi}{6}\tan\frac{5\pi}{4}} C. \frac{\sin\frac{\pi}{6}\cos\frac{5\pi}{4} + \cos\frac{\pi}{6}\sin\frac{5\pi}{4}}{} B. \frac{\tan\frac{\pi}{6} – \tan\frac{5\pi}{4}}{1 + \tan\frac{\pi}{6}\tan\frac{5\pi}{4}} D. Find the exact value of the expression. \tan\left(\frac{\pi}{6} + \frac{5\pi}{4}\right) = \sqrt{7 + 4\sqrt{3}} (Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression. Rationalize all denominators.)

The Correct Answer and Explanation is:

To evaluate

tan⁡(π6+5π4)\tan\left(\frac{\pi}{6} + \frac{5\pi}{4}\right)

we use the tangent sum identity:

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}

Let A=π6A = \frac{\pi}{6} and B=5π4B = \frac{5\pi}{4}. Applying the formula:

tan⁡(π6+5π4)=tan⁡π6+tan⁡5π41−tan⁡π6tan⁡5π4\tan\left(\frac{\pi}{6} + \frac{5\pi}{4}\right) = \frac{\tan\frac{\pi}{6} + \tan\frac{5\pi}{4}}{1 – \tan\frac{\pi}{6} \tan\frac{5\pi}{4}}

From the unit circle:

• tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}
• tan⁡5π4=1\tan\frac{5\pi}{4} = 1

Substituting:

=13+11−(13⋅1)=1+33−1= \frac{\frac{1}{\sqrt{3}} + 1}{1 – \left(\frac{1}{\sqrt{3}} \cdot 1\right)} = \frac{1 + \sqrt{3}}{\sqrt{3} – 1}

Rationalizing the denominator:

1+33−1⋅3+13+1=(1+3)(3+1)(3)2−12\frac{1 + \sqrt{3}}{\sqrt{3} – 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3})^2 – 1^2}

=(1⋅3+1⋅1+3⋅3+3⋅1)3−1=3+1+3+32=4+232=2+3= \frac{(1 \cdot \sqrt{3} + 1 \cdot 1 + \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1)}{3 – 1} = \frac{\sqrt{3} + 1 + 3 + \sqrt{3}}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}

Hence, the exact value is 2+3\boxed{2 + \sqrt{3}}.

This confirms that the correct choice for rewriting the expression is:

tan⁡π6+tan⁡5π41−tan⁡π6tan⁡5π4\frac{\tan\frac{\pi}{6} + \tan\frac{5\pi}{4}}{1 – \tan\frac{\pi}{6} \tan\frac{5\pi}{4}}

which corresponds to option A.

The final numerical expression 2+32 + \sqrt{3} is not equal to 7+43\sqrt{7 + 4\sqrt{3}}, despite appearances. Squaring both:

• (2+3)2=4+43+3=7+43(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

Thus, 2+3=7+432 + \sqrt{3} = \sqrt{7 + 4\sqrt{3}}, confirming the equality.

In summary:

• Use the tangent sum formula
• Plug in known tangent values
• Rationalize carefully
• Confirm numerical identity by squaring

Beautifully satisfying, isn’t it?