First slide

Trigonometric Identities

Question

If $\frac{\sin x}{a} = \frac{\cos x}{b} = \frac{\tan x}{c} = k$ then $bc + \frac{1}{ck} + \frac{ak}{1 + bk}$ is equal to

Moderate

A

$k (a + \frac{1}{a})$
B

$\frac{1}{k} (a + \frac{1}{a})$
C

$\frac{1}{k^{2}}$
D

$\frac{a}{k}$

Solution

$\begin{aligned} bc + \frac{1}{ck} + \frac{ak}{1 + bk} = \frac{\cos xtan x}{k^{2}} + \frac{1}{\tan x} + \frac{\sin x}{1 + \cos x} \\ = \frac{\sin x}{k^{2}} + \frac{\cos x (1 + \cos x) + \sin^{2} x}{\sin x (1 + \cos x)} \\ = \frac{a}{k} + \frac{\cos x (1 + \cos x) + (1 - \cos^{2} x)}{\sin x (1 + \cos x)} \\ = \frac{a}{k} + \frac{1}{\sin x} \\ = \frac{a}{k} + \frac{1}{ak} \end{aligned}$

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