Expert answers to "If 2tan⁡A=3tan⁡B then prove thattan⁡(A−B)=sin⁡2B5−cos⁡2B"

Kaylee Torres

Answered question

2022-04-24

2 \tan A = 3 \tan B

then prove that

\tan (A - B) = \frac{\sin 2 B}{5 - \cos 2 B}

Spencer Murillo

Beginner2022-04-25Added 9 answers

Notice, here is another method:

RHS = \frac{\sin 2 B}{5 - \cos 2 B}

= \frac{\frac{2 \tan B}{1 + \tan^{2} B}}{5 - \frac{1 - \tan^{2} B}{1 + \tan^{2} A}}

= \frac{2 \tan B}{4 + 6 \tan^{2} B}

= \frac{\tan B}{2 + 3 \tan^{2} B}

= \frac{(3 \tan B) - 2 \tan B}{2 + (3 \tan B) \tan B}

setting

3 \tan B = 2 \tan A

= \frac{2 \tan A - 2 \tan B}{2 + 2 \tan A \tan B}

= \frac{\tan A - \tan B}{1 + \tan A \tan B}

= \tan (A - B) = LHS

tswe0uk

Beginner2022-04-26Added 19 answers

We have

\tan (A - B) = \frac{\tan (A) - \tan (B)}{1 + \tan (A) \tan (B)} = \frac{\frac{\tan (B)}{2}}{1 + \frac{3}{2} \tan^{2} (B)}

= \frac{\tan (B)}{2 + 3 \tan^{2} (B)} = \frac{\frac{\sin (B)}{\cos (B)}}{2 + 3 \cdot \frac{\sin^{2} (B)}{\cos^{2} (B)}}

Hence, we have

\tan (A - B) = \frac{\sin (B) \cos (B)}{2 \cos^{2} (B) + 3 \sin^{2} (B)} = \frac{\sin (2 B)}{4 + 2 \sin^{2} (B)}

= \frac{\sin (2 B)}{5 - \cos (2 B)}

where we made use of

\sin (2 B) = 2 \sin (B) \cos (B), \cos (2 B) = 1 - 2 \sin^{2} (B)

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