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Question

Mathematics Question on Properties of Inverse Trigonometric Functions

To derive the tangent formula, the following steps are given: 1. tan(A+B)=sinAcosBcosAcosB+cosAsinBcosAcosBcosAcosBsinAsinB+sinAsinBcosAcosB\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(A+B)=sin(A+B)cos(A+B)\tan\left(A + B\right) = \frac{\sin\left(A + B\right)}{\cos\left(A + B\right)} 3. tan(A+B)=sinAcosB+cosAsinBcosAcosBsinAsinB\tan\left(A + B\right) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} 4.tan(A+B)=tanA+tanB1tanAtanB\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

2, 3, 1, 4

Explanation

Solution

Tangent formula is derived as follows tan(A+B)=sin(A+B)cos(A+B)\tan \left(A +B\right) = \frac{\sin\left(A+B\right)}{\cos\left(A +B\right)} =sinAcosB+cosAsinBcosAcosBsinAsinB= \frac{\sin A \cos B + \cos A \sin B}{ \cos A \cos B - \sin A \sin B} =sinAcosBcosAcosB+cosAsinBcosAcosBcosAcosBcosAcosBsinAsinBcosAcosB=tanA+tanB1tanAtanB= \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.