Question

The expression $(\tan^4x + 2\tan^2x + 1) \times \frac{1}{1+\tan^2x}$ when $x = \frac{\pi}{12}$ can be equal to

• A. $4(2 - \sqrt{3})$
• B. $4(\sqrt{2} + 1)$
• C. $16\cos^2\frac{\pi}{12}$
• D. $16\sin^2\frac{\pi}{12}$

Answer 1

Answer: (A), (D)

A, D

Answer 2

The given expression is $(\tan^4x + 2\tan^2x + 1) \times \frac{1}{1+\tan^2x}$.

1. Simplify the first term: The term $(\tan^4x + 2\tan^2x + 1)$ is a perfect square. Let $A = \tan^2x$. Then the expression becomes $(A^2 + 2A + 1) = (A+1)^2 = (\tan^2x + 1)^2$.

2. Use trigonometric identity: Recall the identity $1+\tan^2x = \sec^2x$. So, the first term becomes $(\sec^2x)^2 = \sec^4x$. The second term is $\frac{1}{1+\tan^2x} = \frac{1}{\sec^2x}$.

3. Combine the simplified terms: The expression simplifies to $\sec^4x \times \frac{1}{\sec^2x} = \sec^2x$.

4. Evaluate at $x = \frac{\pi}{12}$: We need to find the value of $\sec^2(\frac{\pi}{12})$. This is equivalent to $\frac{1}{\cos^2(\frac{\pi}{12})}$. First, calculate $\cos(\frac{\pi}{12})$ (which is $\cos(15^\circ)$): $\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$ $= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

5. Calculate $\cos^2(\frac{\pi}{12})$: $\cos^2(\frac{\pi}{12}) = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2 = \frac{(\sqrt{6})^2 + (\sqrt{2})^2 + 2\sqrt{6}\sqrt{2}}{16}$ $= \frac{6 + 2 + 2\sqrt{12}}{16} = \frac{8 + 4\sqrt{3}}{16} = \frac{4(2 + \sqrt{3})}{16} = \frac{2 + \sqrt{3}}{4}$.

6. Calculate $\sec^2(\frac{\pi}{12})$: $\sec^2(\frac{\pi}{12}) = \frac{1}{\cos^2(\frac{\pi}{12})} = \frac{1}{\frac{2 + \sqrt{3}}{4}} = \frac{4}{2 + \sqrt{3}}$. Rationalize the denominator: $\frac{4}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3})}{2^2 - (\sqrt{3})^2} = \frac{4(2 - \sqrt{3})}{4 - 3} = \frac{4(2 - \sqrt{3})}{1} = 4(2 - \sqrt{3})$.

Thus, $\sec^2(\frac{\pi}{12}) = 4(2 - \sqrt{3})$.

Now let's verify the options: (A) $4(2 - \sqrt{3})$ - Matches our result.

(C) $16\cos^2\frac{\pi}{12} = 16 \times \frac{2 + \sqrt{3}}{4} = 4(2 + \sqrt{3})$. This is not $4(2 - \sqrt{3})$.

(D) $16\sin^2\frac{\pi}{12}$. Let's calculate $\sin^2(\frac{\pi}{12})$. $\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ $= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}$. $\sin^2(\frac{\pi}{12}) = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \frac{(\sqrt{6})^2 + (\sqrt{2})^2 - 2\sqrt{6}\sqrt{2}}{16}$ $= \frac{6 + 2 - 2\sqrt{12}}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{4(2 - \sqrt{3})}{16} = \frac{2 - \sqrt{3}}{4}$. So, $16\sin^2\frac{\pi}{12} = 16 \times \frac{2 - \sqrt{3}}{4} = 4(2 - \sqrt{3})$. This also matches our result.

Since the question states "can be equal to", both (A) and (D) are correct.