Question

Help please: The equation tan(x+π/6) is equal to A. √3 tanx +1/√3– tan x B. tanx - √3/1+ √3tan x c. tanx+ √3 /1- √3 tanx D. √3 tanx - 1 /√3 +tan x

Answer 1

A.

`\frac{\sqrt[]{3}\tan x+1}{\sqrt[]{3}-\tan x}`

Explanation:

We have the following equation:

`tan\mleft(x+(\pi)/(6)\mright)`

We solve to calculate its equivalence:

`\begin{gathered} \tan x=(\sin x)/(\cos x) \\ \text{therefore:} \\ (\sin (x+(\pi)/(6)))/(\cos (x+(\pi)/(6))) \\ \sin (s+t)=\sin s\cdot\cos t+\cos s\cdot\sin t \\ \cos (s+t)=\cos s\cos t-\sin s\cdot\sin t \\ \text{ replacing:} \\ (\sin x\cdot\cos(\pi)/(6)+\cos x\cdot\sin(\pi)/(6))/(\cos x\cos(\pi)/(6)-\sin x\cdot\sin(\pi)/(6))=\frac{\frac{\sqrt[]{3}}{2}\sin x+(1)/(2)\cos x}{\frac{\sqrt[]{3}}{2}\cos x-(1)/(2)\sin x} \\ \cos x=(1)/(\sec x) \\ \frac{\frac{\sqrt[]{3}}{2}\sin x+(1)/(2)(1)/(\sec x)}{\frac{\sqrt[]{3}}{2}(1)/(\sec x)-(1)/(2)\sin x}=\frac{\frac{1+\sqrt[]{3}\cdot\sin x\cdot\sec x}{2\sec x}}{\frac{\sqrt[]{3}-\sin x\cdot\sec x}{2\sec x}}=\frac{1+\sqrt[]{3}\cdot\sin x\cdot\sec x}{\sqrt[]{3}-\sin x\cdot\sec x} \\ \sin x\cdot\sec x=\tan x \\ \text{ replacing} \\ \frac{1+\sqrt[]{3}\cdot\tan x}{\sqrt[]{3}-\tan x}=\frac{\sqrt[]{3}\tan x+1}{\sqrt[]{3}-\tan x} \end{gathered}`