Trigonometric Equation Solution: Finding tan(15θ)

The problem requires finding the value of $\tan 15\theta$ given the condition $\cot 3\theta \cot 6\theta = 1$. We will use trigonometric identities to solve for $\theta$ and then evaluate $\tan 15\theta$.

Key Trigonometric Identity

The condition $\cot A \cot B = 1$ can be rewritten as:

$ \frac{\cos A \cos B}{\sin A \sin B} = 1 $

$ \cos A \cos B - \sin A \sin B = 0 $

This corresponds to the identity for $\cos(A+B)$:

$ \cos(A+B) = 0 $

The general solution for $\cos x = 0$ is $x = 90^\circ + n \times 180^\circ$, where $n$ is an integer.

Applying the Identity

In this problem, $A = 3\theta$ and $B = 6\theta$. Applying the identity derived above:

• $3\theta + 6\theta = 90^\circ + n \times 180^\circ$
• $9\theta = 90^\circ + n \times 180^\circ$

Now, solve for $\theta$ by dividing by 9:

$ \theta = \frac{90^\circ}{9} + n \times \frac{180^\circ}{9} $

$ \theta = 10^\circ + n \times 20^\circ $

Calculating tan(15θ)

We need to find the value of $\tan 15\theta$. Substitute the expression for $\theta$:

$ 15\theta = 15 \times (10^\circ + n \times 20^\circ) $

$ 15\theta = 150^\circ + n \times 300^\circ $

Now, evaluate $\tan 15\theta$ using this expression. Let's test a simple case, for $n=0$:

• If $n=0$, then $15\theta = 150^\circ$.
• $\tan(150^\circ) = \tan(180^\circ - 30^\circ)$
• Using the identity $\tan(180^\circ - x) = -\tan x$:
• $\tan(150^\circ) = -\tan(30^\circ)$
• $ -\tan(30^\circ) = -\frac{1}{\sqrt{3}} $

This value matches one of the options.

Conclusion

Given $\cot 3\theta \cot 6\theta = 1$, we found $9\theta = 90^\circ + n \times 180^\circ$. Evaluating $\tan 15\theta$ for $n=0$ yields $\tan 150^\circ$, which equals $-\frac{1}{\sqrt{3}}$.