Trigonometry & Inverse Trigonometry
Trigonometric Equations and Identities
Grade 11
Question:
<p>If \(x\), \(y\) and \(z\) are real numbers that satisfy the three equations<br>\[\begin{cases} \tan(x)+\tan(y)+\tan(z) = 6-(\cot(x)+\cot(y)+\cot(z))\\ \tan^2(x)+\tan^2(y)+\tan^2(z) = 6-(\cot^2(x)+\cot^2(y)+\cot^2(z))\\ \tan^3(x)+\tan^3(y)+\tan^3(z) = 6-(\cot^3(x)+\cot^3(y)+\cot^3(z)) \end{cases}\]<br>Find the value of the expression \(\left(\dfrac{\tan(x)}{\tan(y)}+\dfrac{\tan(y)}{\tan(z)}+\dfrac{\tan(z)}{\tan(x)}+3\tan(x)\tan(y)\tan(z)\right)\).</p>
Step-by-Step Solution
Key Concept: Recognize that if we substitute $t_i = \tan(x_i)$, the three equations establish a symmetric relationship between $t_i$ and $1/t_i$ values. The key is to observe that each equation has the form $S_n(t) = 6 - S_n(1/t)$, suggesting that the tangent values satisfy a special constraint.
<p><strong>Step 1:</strong> Let $a = \tan(x)$, $b = \tan(y)$, $c = \tan(z)$. The three equations become:</p><p>$$a + b + c = 6 - \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$</p><p>$$a^2 + b^2 + c^2 = 6 - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$</p><p>$$a^3 + b^3 + c^3 = 6 - \left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right)$$</p><p><strong>Step 2:</strong> Rewrite the first equation:</p><p>$$a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 6$$</p><p>$$a + b + c + \frac{ab + bc + ca}{abc} = 6$$</p><p><strong>Step 3:</strong> Note that $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2 - 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right)$</p><p><strong>Step 4:</strong> From the symmetry of the equations and testing: if $a = b = c = k$, then:</p><p>$$3k = 6 - \frac{3}{k} \implies 3k + \frac{3}{k} = 6 \implies k + \frac{1}{k} = 2$$</p><p>This gives $k^2 - 2k + 1 = 0$, so $k = 1$.</p><p><strong>Step 5:</strong> Verify: If $\tan(x) = \tan(y) = \tan(z) = 1$:</p><p>- Equation 1: $1 + 1 + 1 = 6 - (1 + 1 + 1) = 3$ ✓</p><p>- Equation 2: $1 + 1 + 1 = 6 - (1 + 1 + 1) = 3$ ✓</p><p>- Equation 3: $1 + 1 + 1 = 6 - (1 + 1 + 1) = 3$ ✓</p><p><strong>Step 6:</strong> Calculate the required expression with $a = b = c = 1$:</p><p>$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + 3abc = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + 3(1)(1)(1)$$</p><p>$$= 1 + 1 + 1 + 3 = 6$$</p><p><strong>∴ Answer: 6</strong></p>
Correct Answer: 6