Question

use identities to write the following expression as a function of z alone. tan (60° + z) choose the correct answer below. a. \frac{sqrt{3}-\tan z}{1+sqrt{3} \tan z} b. \frac{sqrt{3}+\tan z}{1-sqrt{3} \tan z} c. \frac{sqrt{3}-2 \tan z}{sqrt{3}+2 \tan z} d. \frac{sqrt{3}+2 \tan z}{sqrt{3}-2 \tan z}

Explanation:

Step1: Apply the tangent addition formula

The tangent addition formula is \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\). Here \(A = 60^{\circ}\) and \(B=z\). We know that \(\tan60^{\circ}=\sqrt{3}\).
Substituting \(A = 60^{\circ}\) and \(B = z\) into the formula, we get \(\tan(60^{\circ}+z)=\frac{\tan60^{\circ}+\tan z}{1 - \tan60^{\circ}\tan z}\).

Step2: Substitute the value of \(\tan60^{\circ}\)

Since \(\tan60^{\circ}=\sqrt{3}\), the expression becomes \(\frac{\sqrt{3}+\tan z}{1-\sqrt{3}\tan z}\).

Answer:

B. \(\frac{\sqrt{3}+\tan z}{1 - \sqrt{3}\tan z}\)