Let k= tan -1(( sin (π/18)+ sin (2 π/18)/ cos (π)18+ cos (2 π)18), then k equals

Question

Let k= tan -1(( sin (π/18)+ sin (2 π/18)/ cos (π)18+ cos (2 π)18), then k equals (A) (π/8) (B) (π/12) (C) (π/15) (D) (π/4)

Tardigrade Admin · Accepted Answer

θ= tan -1(( sin (π/18)+ sin (2 π/18)/ cos (π)18+ cos (2 π)18)= tan /-1)((2 sin (3 π/36) cos (π/36)/2 cos (3 π)36 cos (π)36)= tan /-1)( tan (3 π/36)) θ=(3 π/36) radian =(π/12)

Q. Let $k = tan^{- 1} (\frac{s i n \frac{π}{18} + s i n \frac{2 π}{18}}{c o s \frac{π}{18} + c o s \frac{2 π}{18}})$ , then $k$ equals

$\frac{π}{8}$

$\frac{π}{12}$

$\frac{π}{15}$

$\frac{π}{4}$

Q. Let k=tan−1(cos18π​+cos182π​sin18π​+sin182π​​), then k equals

8π​

12π​

15π​

4π​

θ=tan−1(cos18π​+cos182π​sin18π​+sin182π​​)=tan−1(2cos363π​cos36π​2sin363π​cos36π​​)=tan−1(tan363π​) θ=363π​ radian =12π​

Q. Let $k = tan^{- 1} (\frac{s i n \frac{π}{18} + s i n \frac{2 π}{18}}{c o s \frac{π}{18} + c o s \frac{2 π}{18}})$ , then $k$ equals

$\frac{π}{8}$

$\frac{π}{12}$

$\frac{π}{15}$

$\frac{π}{4}$