Question

To derive the tangent formula, the following steps are given:
1. $\tan\left(A + B\right) = \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\sin A \sin B} + \frac{\sin A \sin B}{\cos A \cos B}}$
2. $\tan\left(A + B\right) = \frac{\sin\left(A + B\right)}{\cos\left(A + B\right)} $
3. $\tan\left(A + B\right) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}$
4.$ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1- \tan A \tan B} $
Their correct and proper sequential form to derive the formula is:

• A. 2, 4, 3, 1
• B. 1, 2, 3, 4
• C. 1, 4, 2, 3
• D. 2, 3, 1, 4

Answer 1

Answer: (D)

Tangent formula is derived as follows
$\tan \left(A +B\right) = \frac{\sin\left(A+B\right)}{\cos\left(A +B\right)} $
$= \frac{\sin A \cos B + \cos A \sin B}{ \cos A \cos B - \sin A \sin B}$
$= \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} - \frac{\sin A \sin B}{\cos A \cos B}} = \frac{\tan A + \tan B}{1- \tan A \tan B}$
Correct and proper sequential form to derive the formula is 2, 3, 1, 4.