First slide

Compound angles

Question

If $\frac{\sin x}{\sin y} = \frac{1}{2}, \frac{\cos x}{\cos y} = \frac{3}{2}, where x, y \in (0, \frac{π}{2}),$ then the value of tan(x + y) is equal to

Moderate

A

$\sqrt{13}$
B

$\sqrt{14}$
C

$\sqrt{17}$
D

$\sqrt{15}$

Solution

$\begin{array}{l} \frac{\sin x}{\sin y} = \frac{1}{2}, \frac{\cos x}{\cos y} = \frac{3}{2} \\ \Rightarrow \frac{\tan x}{\tan y} = \frac{1}{3} \\ \Rightarrow \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan xtan y} = \frac{4 \tan x}{1 - 3 \tan^{2} x} \end{array}$
$\begin{array}{l} Also \sin y = 2 \sin x, \cos y = \frac{2}{3} \cos x \\ or \sin^{2} y + \cos^{2} y = 4 \sin^{2} x + \frac{4 \cos^{2} x}{9} = 1 \\ or 36 \tan^{2} x + 4 = 9 \sec^{2} x = 9 (1 + \tan^{2} x) \\ or 27 \tan^{2} x = 5 \\ or \tan x = \frac{\sqrt{5}}{3 \sqrt{3}} \\ \Rightarrow \tan (x + y) = \frac{\frac{4 \sqrt{5}}{3 \sqrt{3}}}{1 - \frac{15}{27}} = \frac{4 \sqrt{5} \times 27}{12 \times 3 \sqrt{3}} = \sqrt{15} \end{array}$

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