The value of tan (13 π/12) is

Question

The value of tan (13 π/12) is (A) 2 (B) 2+√3 (C) 2-√3 (D) √3

Tardigrade Admin · Accepted Answer

We have, tan (13 π/12) = tan (π+(π/12))= tan (π/12)= tan ((π/4)-(π/6)) =( tan (π/4)- tan (π/6)/1+ tan (π)4 tan (π)6=(1-(1/√3)/1+(1)√3)=(√3-1/√3+1)=2-√/3)

Q. The value of $tan \frac{13 π}{12}$ is

2

$2 + 3$

$2 - 3$

$3$

We have,
$tan \frac{13 π}{12} = tan (π + \frac{π}{12}) = tan \frac{π}{12} = tan (\frac{π}{4} - \frac{π}{6})$
$= \frac{t a n \frac{π}{4} - t a n \frac{π}{6}}{1 + t a n \frac{π}{4} t a n \frac{π}{6}} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{3 - 1}{3 + 1} = 2 - 3$

Q. The value of tan1213π​ is

2

2+3​

2−3​

3​

We have, tan1213π​=tan(π+12π​)=tan12π​=tan(4π​−6π​) =1+tan4π​tan6π​tan4π​−tan6π​​=1+3​1​1−3​1​​=3​+13​−1​=2−3​

Q. The value of $tan \frac{13 π}{12}$ is

$2 + 3$

$2 - 3$

$3$

We have,
$tan \frac{13 π}{12} = tan (π + \frac{π}{12}) = tan \frac{π}{12} = tan (\frac{π}{4} - \frac{π}{6})$
$= \frac{t a n \frac{π}{4} - t a n \frac{π}{6}}{1 + t a n \frac{π}{4} t a n \frac{π}{6}} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{3 - 1}{3 + 1} = 2 - 3$