Question

doing this and writing the results in terms of sine and cosine yields
(simplify your answers.)
the previous expression equals \\( \frac { \tan x - \tan y } { \tan x + \tan y } \\) by which of the following?
a. factoring \\( \cos x \\) from the terms in the numerator
b. the quotient identity for tangent
c. factoring and applying the sum identity for tangent
d. a reciprocal angle identity
e. a negative angle identity
f. a pythagorean identity

Explanation:

The quotient identity for tangent is \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). When we write \(\tan x\) as \(\frac{\sin x}{\cos x}\) and \(\tan y\) as \(\frac{\sin y}{\cos y}\) and then simplify the expression \(\frac{\tan x - \tan y}{\tan x+\tan y}\), we are using the quotient identity for tangent.

• Option A: Factoring \(\cos x\) from the terms in the numerator is not relevant here as the transformation is based on the definition of tangent.
• Option C: The sum identity for tangent is \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\), which is not used in this simplification.
• Option D: Reciprocal angle identities are for example \(\csc\theta=\frac{1}{\sin\theta}\), \(\sec\theta=\frac{1}{\cos\theta}\), \(\cot\theta=\frac{1}{\tan\theta}\), not applicable here.
• Option E: Negative - angle identities are \(\sin(-\theta)=-\sin\theta\), \(\cos(-\theta)=\cos\theta\), \(\tan(-\theta)=-\tan\theta\), not relevant.
• Option F: Pythagorean identities are \(\sin^{2}\theta+\cos^{2}\theta = 1\), \(1+\tan^{2}\theta=\sec^{2}\theta\), \(1+\cot^{2}\theta=\csc^{2}\theta\), not used here.

Answer:

B. The quotient identity for tangent