First slide

Trigonometric equations

Question

If $\frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} = A (\tan B θ - \tan C θ)$ , then $(\frac{B + C}{A}) = (A > 0, θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$

Moderate

A

54
B

55
C

56
D

57

Solution

$W e h a v e \frac{\sin θ}{\cos 3 θ} = \frac{1}{2} (\frac{\sin 2 θ}{\cos θ \cos 3 θ}) \frac{1}{2} \frac{\sin (3 θ - θ)}{\cos θ \cos 3 θ} = \frac{1}{2} \frac{\sin 3 θ \cos θ - \cos 3 θ \sin θ}{\cos θ \cos 3 θ} \frac{\sin θ}{\cos 3 θ} = \frac{1}{2} (\tan 3 θ - \tan θ) - - - - - (1)$
$\begin{array}{l} S i m i l a r l y \frac{\sin 3 θ}{\cos 9 θ} = \frac{1}{2} (\tan 9 θ - \tan 3 θ) - - - - (2) \\ a n d \frac{\sin 9 θ}{\cos 27 θ} = \frac{1}{2} (\tan 27 θ - \tan 9 θ) - - - - (3) \end{array}$
adding (1) ,(2) ,(3)
$∴ \frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} = \frac{1}{2} (\tan 27 θ - \tan θ)$ $= A (\tan B θ - \tan C θ)$
$∴ (\frac{B + C}{A}) = \frac{27 + 1}{\frac{1}{2}} = 56$

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