Question

$\tan 10^\circ = \tan 20^\circ \cdot \tan 30^\circ \cdot \tan 40^\circ$.

Answer 1

True

Answer 2

To verify the given statement: $\tan 10^\circ = \tan 20^\circ \cdot \tan 30^\circ \cdot \tan 40^\circ$.

We will evaluate the Right Hand Side (RHS) of the equation and check if it equals the Left Hand Side (LHS).

RHS = $\tan 20^\circ \cdot \tan 30^\circ \cdot \tan 40^\circ$

We know the exact value of $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Substitute this value into the RHS: RHS = $\tan 20^\circ \cdot \frac{1}{\sqrt{3}} \cdot \tan 40^\circ$ RHS = $\frac{1}{\sqrt{3}} (\tan 20^\circ \tan 40^\circ)$

Now, we use a standard trigonometric identity for products of tangents: The identity is: $\tan \theta \tan (60^\circ - \theta) \tan (60^\circ + \theta) = \tan (3\theta)$

Let's prove this identity for completeness: LHS = $\tan \theta \tan (60^\circ - \theta) \tan (60^\circ + \theta)$ Using the formula $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ and $\tan 60^\circ = \sqrt{3}$: $\tan (60^\circ - \theta) = \frac{\tan 60^\circ - \tan \theta}{1 + \tan 60^\circ \tan \theta} = \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta}$ $\tan (60^\circ + \theta) = \frac{\tan 60^\circ + \tan \theta}{1 - \tan 60^\circ \tan \theta} = \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta}$

Substitute these into the LHS: LHS = $\tan \theta \cdot \left( \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta} \right) \cdot \left( \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta} \right)$ LHS = $\tan \theta \cdot \frac{(\sqrt{3})^2 - (\tan \theta)^2}{1^2 - (\sqrt{3} \tan \theta)^2}$ LHS = $\tan \theta \cdot \frac{3 - \tan^2 \theta}{1 - 3 \tan^2 \theta}$

This is the formula for $\tan(3\theta)$: $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan\theta \frac{3 - \tan^2\theta}{1 - 3\tan^2\theta}$ Thus, the identity $\tan \theta \tan (60^\circ - \theta) \tan (60^\circ + \theta) = \tan (3\theta)$ is proven.

Now, let's apply this identity to the expression $\tan 20^\circ \tan 40^\circ$. Let $\theta = 20^\circ$. Then $60^\circ - \theta = 60^\circ - 20^\circ = 40^\circ$. And $60^\circ + \theta = 60^\circ + 20^\circ = 80^\circ$.

Using the identity: $\tan 20^\circ \tan (60^\circ - 20^\circ) \tan (60^\circ + 20^\circ) = \tan (3 \cdot 20^\circ)$ $\tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 60^\circ$

From this, we can express the product $\tan 20^\circ \tan 40^\circ$: $\tan 20^\circ \tan 40^\circ = \frac{\tan 60^\circ}{\tan 80^\circ}$

Now substitute this back into the RHS of the original equation: RHS = $\frac{1}{\sqrt{3}} (\tan 20^\circ \tan 40^\circ)$ RHS = $\frac{1}{\sqrt{3}} \cdot \frac{\tan 60^\circ}{\tan 80^\circ}$

We know $\tan 60^\circ = \sqrt{3}$. RHS = $\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\tan 80^\circ}$ RHS = $\frac{1}{\tan 80^\circ}$

We also know that $\frac{1}{\tan x} = \cot x$. So, $\frac{1}{\tan 80^\circ} = \cot 80^\circ$. And, using the complementary angle identity $\cot x = \tan (90^\circ - x)$: $\cot 80^\circ = \tan (90^\circ - 80^\circ) = \tan 10^\circ$.

So, RHS = $\tan 10^\circ$.

The Left Hand Side (LHS) of the given equation is $\tan 10^\circ$. Since LHS = $\tan 10^\circ$ and RHS = $\tan 10^\circ$, we have LHS = RHS.

Therefore, the given statement is true.