First slide

Multiple and sub- multiple Angles

Question

$15 [\tan 2 θ + \sin 2 θ] + 8 = 0$ if

Moderate

A

$\tan θ = 1 / 2$
B

$\sin θ = 1 / 4$
C

$\tan θ = 2$
D

$\cos θ = 1 / 5$

Solution

We have $\tan 2 θ + \sin 2 θ = - 8 / 15$
$\Rightarrow \frac{2 \tan θ}{1 - \tan^{2} θ} + \frac{2 \tan θ}{1 + \tan^{2} θ} = \frac{- 8}{15}$
$\begin{array}{l} \Rightarrow 15 \times 4 \tan θ + 8 (1 - \tan^{4} θ) = 0 \\ \Rightarrow 8 \tan^{4} θ + 60 \tan θ - 8 = 0 \end{array}$
which is satisfied if $\tan θ = 2$

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