Question

if sinx/a=cosx/b=tanx/c=k,then bc+1/ck+ak/1+bk is equal to

Answer 1

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Answer 2

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


Solution :
Steps and Understanding :
1) According to the question, we get
$ak = \sin(x) \\ bk = \cos(x) \\ ck = \tan(x) \\ bc {k}^{2} = \sin(x)$

2) So,
$bc \: + \frac{1}{ck} + \frac{ak}{1 + bk}$
=>
$\frac{ \sin(x) }{ {k}^{2} } + \frac{1}{ \tan(x) } + \frac{ \sin(x) }{1 + \cos(x) } \\ = > \frac{a}{k} + \frac{ \cos(x) }{ \sin(x) } + \frac{ 1 - \cos(x) }{ \sin(x) } \\ = > \frac{a}{k} + \frac{ \cos(x) }{ \sin(x) } + \frac{1}{ \sin(x) } - \frac{ \cos(x) }{ \sin(x) } \\ = > \frac{a}{k} + \frac{1}{ \sin(x) } \\ = > \frac{a}{k} + \frac{1}{ak} \\ = > \frac{1}{k} (a + \frac{1}{a} )$

3)So, Simplified Form of given expression is

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