NTA Abhyas JEE Main2020MathematicsComplex NumberPractice
For the equation 1 - i x 1 + i x = s i n π 7 - i c o s π 7 , if x = c o t k π 28 , then the value of k can be (where i 2 = - 1 )
Options
- A1
- B3
- C5
- D9
Correct answer
D. 9
Step-by-step solution
Applying Componendo and Dividendo, we get, 1 - i x + 1 + i x 1 - i x - 1 + i x = s i n π 7 - i c o s π 7 + 1 s i n π 7 - i c o s π 7 - 1 2 - 2 i x = 1 + c o s 5 π 14 - i s i n 5 π 14 - 1 + c o s 5 π 14 - i s i n 5 π 14 - i x = - 2 s i n 2 5 π 28 - i 2 s i n 5 π 28 c o s 5 π 28 2 c o s 2 5 π 28 - i 2 s i n 5 π 28 c o s 5 π 28 x = 2 s i n 5 π 28 s i n 5 π 28 + i c o s 5 π 28 2 c o s 5 π 28 c o s 5 π 28 - i s i n 5 π 28 × i ⇒ x = t a n 5 π 28 = c o t 9 π 28