First slide

Trigonometric equations

Question

If $3 \tan (θ - 15^{\circ}) = \tan (θ + 15^{\circ}), then θ$ is equal to

Moderate

A

$nπ + \frac{π}{4}, n \in Z$
B

$nπ + \frac{π}{8}, n \in Z$
C

$nπ + \frac{π}{3}, n \in Z$
D

none of these

Solution

$\begin{array}{l} let A = θ + 15^{\circ}, B = θ - 15^{\circ} \\ \Rightarrow A + B = 2 θ and A - B = 30^{\circ} \\ Now, \frac{\tan A}{\tan B} = \frac{3}{1} \\ or \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{3 + 1}{3 - 1} \end{array}$
(applying componendo and dividendo rule)
$\begin{array}{l} or \frac{\sin (A + B)}{\sin (A - B)} = 2 \\ \Rightarrow \sin 2 θ = 2 \sin 30^{\circ} = 1 \\ \Rightarrow 2 θ = 2 nπ + \frac{π}{2} or θ = nπ + \frac{π}{4}, n \in Z \end{array}$

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