(c) (4cos^2 (9 deg) - 3)(4cos^2 (27 deg) - 3) = tan 9 deg

1. Triple-Angle Identity Refresher

For any angle $\theta$:

Divide both sides by $\cos\theta \,(\neq 0)$ to get a very handy form:

2. Apply the Identity to Each Factor

• First factor ($\theta = 9°$): $$4\cos^2(9°) - 3 = \frac{\cos(27°)}{\cos(9°)}$$
• Second factor ($\theta = 27°$): $$4\cos^2(27°) - 3 = \frac{\cos(81°)}{\cos(27°)}$$

3. Multiply the Two Factors

Notice $\cos 27°$ cancels out:

4. Convert $\cos 81°$

Because $81° = 90° - 9°$ we know:

Hence

Exactly what we had to show!

Think of Trigonometry Like Lego Blocks 🧩

1. Big Lego Rule (Triple-Angle Rule)
   If you stack three small angles to make a bigger one, their cosines follow a rule: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$

2. What We Are Asked To Show
   We have two funny blocks:

   • Block-A → $4\cos^2(9°)-3$
   • Block-B → $4\cos^2(27°)-3$ We must prove that putting Block-A and Block-B together (multiplying) magically gives $\tan(9°)$.

3. Idea in One Line
   Turn each block into a simpler form using the Big Lego Rule, cancel the matching pieces, and you will be left with sine over cosine, which is exactly tan.

That’s all! 🙂

Step-by-Step Solution

1. Use the triple-angle identity

   $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$

   Divide both sides by $\cos\theta$:

   $4\cos^2\theta - 3 = \frac{\cos(3\theta)}{\cos\theta}$

2. Rewrite each factor

   • For $\theta = 9°$:

     $4\cos^2 9° - 3 = \frac{\cos 27°}{\cos 9°}$

   • For $\theta = 27°$:

     $4\cos^2 27° - 3 = \frac{\cos 81°}{\cos 27°}$

3. Multiply the factors

   $$\left(4\cos^2 9° - 3\right)\left(4\cos^2 27° - 3\right) = \left(\frac{\cos 27°}{\cos 9°}\right)\left(\frac{\cos 81°}{\cos 27°}\right)$$

   Cancel $\cos 27°$:

   $= \frac{\cos 81°}{\cos 9°}$

4. Convert $\cos 81°$ to $\sin 9°$

   $\cos 81° = \cos(90° - 9°) = \sin 9°$

   Hence

   $\frac{\cos 81°}{\cos 9°} = \frac{\sin 9°}{\cos 9°} = \tan 9°$

5. Conclusion

   $\boxed{\left(4\cos^2 9° - 3\right)\left(4\cos^2 27° - 3\right) = \tan 9°}$