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Question: If \(\tan A = 2\tan B + \cot B,\) then \(2\tan(A - B) =\)...

If tan⁡A=2tan⁡B+cot⁡B,\tan A = 2\tan B + \cot B, then 2tan⁡(A−B)=2\tan(A - B) =

A

tan⁡B\tan B

B

2tan⁡B2\tan B

C

cot⁡B\cot B

D

2cot⁡B2\cot B

Answer

cot⁡B\cot B

Explanation

Solution

2tan⁡(A−B)=2(tan⁡A−tan⁡B1+tan⁡Atan⁡B)2\tan(A - B) = 2\left( \frac{\tan A - \tan B}{1 + \tan A\tan B} \right)

=2(2tan⁡B+cot⁡B−tan⁡B)1+(2tan⁡B+cot⁡B)tan⁡B=2tan⁡B+cot⁡B2(1+tan⁡2B)= 2\frac{(2\tan B + \cot B - \tan B)}{1 + (2\tan B + \cot B)\tan B} = 2\frac{\tan B + \cot B}{2(1 + \tan^{2}B)}

=cot⁡B(tan⁡2B+1)(1+tan⁡2B)=cot⁡B= \frac{\cot B(\tan^{2}B + 1)}{(1 + \tan^{2}B)} = \cot B.