First slide

Trigonometric transformations

Question

If $\frac{\tan (α + β - γ)}{\tan (α - β + γ)} = \frac{\tan γ}{\tan β}, (β \neq γ)$ then $\sin 2 α + \sin 2 β + \sin 2 γ =$

Moderate

A

0
B

1
C

2
D

½

Solution

$\begin{array}{l} \frac{\tan (α + β - γ)}{\tan (α - β + γ)} = \frac{\tan γ}{\tan β} \\ \Rightarrow \frac{\sin (α + β - γ) \cos (α - β + γ)}{\sin (α - β + γ) \cos (α + β - γ)} = \frac{\sin γcos β}{\sin βcos γ} \end{array}$
Applying componendo and dividendo, we get
$\begin{array}{l} \frac{\sin 2 α}{\sin 2 (β - γ)} = \frac{\sin (γ + β)}{\sin (γ - β)} \\ \Rightarrow \sin 2 (β - γ) \sin (β + γ) + \sin 2 αsin (β - γ) = 0 \\ \Rightarrow \sin (β - γ) (2 \cos (β - γ) \sin (β + γ) + \sin 2 α) = 0 \\ \Rightarrow \sin (β - γ) (\sin 2 α + \sin 2 β + \sin 2 γ) = 0 \\ \Rightarrow \sin 2 α + \sin 2 β + \sin 2 γ = 0 \\ \Rightarrow (as β \neq γ) \end{array}$

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