Question

If a = cos a + i sin a, b = cos b + i sin b,

c = cos g + i sin g and $\frac{a}{b}$+$\frac{b}{c}$+$\frac{c}{a}$= 1, then

cos (a – b) + cos (b – g) + cos (g – a) =

• A. $\frac{3}{2}$
• B. –$\frac{3}{2}$
• C. 0
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Sol. $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ = 1

Ž $\frac{cis\alpha}{cis\beta}$+$\frac{cis\beta}{cis\gamma}$+$\frac{cis\gamma}{cis\alpha}$= 1,

Where cis q represents cos q + i sin q

cis (a – b) + cis (b – g) + cis (g – a) = 1

Equation real parts of both sides

Ž cos (a – b) + cos (b – g) + cos (g – a) = 1

Hence (4) is correct answer.