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) text 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−1cos18π​+cos182π​sin18π​+sin182π​​ then k equals

8π​

12π​

15π​

4π​

θ=tan−1cos18π​+cos182π​sin18π​+sin182π​​ =tan−12cos363π​cos36π​2sin363π​cos36π​​ =tan−1tan363π​ 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

$\frac{π}{8}$

$\frac{π}{12}$

$\frac{π}{15}$

$\frac{π}{4}$