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  4. If ( sin x/ a ) = ( cos x/b) = ( tan x/c) = k , then bc + (1/ck) + (ak/1+bk ) =

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Q. If $\frac{\sin x}{ a } = \frac{\cos x}{b} = \frac{\tan x}{c} = k , $ then $bc + \frac{1}{ck} + \frac{ak}{1+bk } = $

Trigonometric Functions

Solution:

$bc + \frac{1}{ck} + \frac{ak}{ 1+ bk}$
$ = \left(\frac{\cos x}{k}\right)\left(\frac{\tan x}{k}\right) + \frac{1}{\tan x} + \frac{\sin x}{ 1+ \cos x} $
$ = \frac{\sin x}{k^{2}} + \left(\frac{\cos x}{\sin x} + \frac{\sin x}{1+\cos x}\right)$
$ = \frac{1}{k} \left(\frac{\sin x}{k}\right) + \frac{\cos x \left(1+ \cos x\right) + \sin^{2} x}{\sin x \left(1+\cos x\right)} $
$ = \frac{a}{k} + \frac{1+ \cos x}{\sin x \left(1+ \cos x\right)} $
$ = \frac{a}{k} + \frac{1}{\sin x} = \frac{a}{k} + \frac{1}{ak} = \frac{1}{k} \left(a+ \frac{1}{a}\right) $