Question

If 2 tan B + cot B = tan A, prove that 2tan (A-B) = cot B​

Answer 1

See below for proof.

Explanation:

Trigonometric identity

`\boxed{\tan (A \pm B)=(\tan A \pm \tan B)/(1 \mp \tan A \tan B)}`

Using the above trigonometric identity, we can rewrite 2 tan (A - B) as:

`\implies 2 \tan (A-B)=(2(\tan A - \tan B))/(1+\tan A \tan B)`

Given that 2 tan B + cot B = tan A, substitute this into the fraction:

`=(2(2 \tan B + \cot B - \tan B))/(1+(2 \tan B + \cot B) \tan B)`

Simplify:

`=(2(\tan B + \cot B))/(1+2 \tan^2 B + \cot B \tan B)`

`\textsf{Use\;the\;trigonometric\;identity:\;\;$\cot x=(1)/(\tan x)$}`

`=(2\left( \tan B + (1)/(\tan B)\right))/(1+2 \tan^2 B + ( \tan B)/(\tan B))`

Simplify:

`=(2\left((\tan^2B+1)/(\tan B)\right))/(1+2 \tan^2 B +1)`

`=(2\left((\tan^2B+1)/(\tan B)\right))/(2(\tan^2 B +1))`

`=((\tan^2B+1)/(\tan B))/(\tan^2 B +1)`

`\textsf{Apply\:the\:fraction\:rule:\;\; $((a)/(b))/(c)=(a)/(b\cdot c)$}`

`=(\tan^2B+1)/(\tan B(\tan^2 B +1))`

Cancel the common factor (tan²B + 1):

`=(1)/(\tan B)`

`\textsf{Use\:the\:trigonometric\:identity:\quad $(1)/(\tan x)=\cot x$}`

`=\cot B`

`\hrulefill`

As one calculation:

`\begin{aligned}\implies 2\tan (A-B)&=(2(\tan A-\tan B))/(1+\tan A\tan B)\\\\&=(2(2\tan B+\cot B-\tan B))/(1+(2\tan B+\cot B)\tan B)\\\\&=(2(\tan B+\cot B))/(1+2\tan^2 B+\cot B\tan B)\\\\&=(2\left(\tan B+(1)/(\tan B)\right))/(1+2\tan^2 B+(\tan B)/(\tan B))\\\\&=(2\left((\tan^2B+1)/(\tan B)\right))/(1+2\tan^2 B+1)\\\\&=(2\left((\tan^2B+1)/(\tan B)\right))/(2(\tan^2 B+1))\\\\&=((\tan^2B+1)/(\tan B))/(\tan^2 B+1)\\\\\end{aligned}`

`\begin{aligned}&=(\tan^2B+1)/(\tan B(\tan^2 B+1))\\\\&=(1)/(\tan B)\\\\&=\cot B\end{aligned}`