If $ \dfrac{{\sin {\rm{x}}}}{{\rm{a}}} = \dfrac{{\cos {\rm{x}}}}{{\rm{b}}} = \dfrac{{\tan {\rm{x}}}}{{\rm{c}}} = {\rm{k}}$ then ${\rm{bc}} + \dfrac{1}{{{\rm{ck}}}} + \dfrac{{{\rm{ak}}}}{{1 + 6{\rm{k}}}}$ is
A) ${\rm{k}}\left( {{\rm{a}} + \dfrac{1}{{\rm{a}}}} \right)$
B) $\dfrac{{\rm{a}}}{{\rm{k}}} + \dfrac{1}{{{\rm{ak}}}}{\rm{\;}}$
C) $\dfrac{1}{{{{\rm{k}}^2}}}$
D) $\dfrac{{\rm{a}}}{{\rm{k}}}$
Answer
621k+ views
Hint:
First of all we’ll find the values of a, b and c in terms of k and put the values in ${\rm{bc}} + \dfrac{1}{{{\rm{ck}}}} + \dfrac{{{\rm{ak}}}}{{1 + 6{\rm{k}}}}$. Then we’ll simplify further using trigonometric formulas to get the answer.
Complete step by step solution:
$\dfrac{{\sin {\rm{x}}}}{{\rm{a}}} = \dfrac{{\cos {\rm{x}}}}{{\rm{b}}} = \dfrac{{\tan {\rm{x}}}}{{\rm{c}}} = {\rm{k}}$ $\left[ {\dfrac{{\sin {\rm{x}}}}{{\rm{k}}} = {\rm{a}}} \right]$
${\rm{bc}} + \dfrac{1}{{{\rm{ck}}}} + \dfrac{{{\rm{ak}}}}{{1 + {\rm{bk}}}}$
We can write
$ \Rightarrow \dfrac{{\cos {\rm{x}}}}{{\rm{k}}} \times \dfrac{{\tan {\rm{x}}}}{{\rm{k}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{\cos {\rm{x}}}}{{\rm{k}}} \times \dfrac{{\sin {\rm{x}}}}{{\cos {\rm{x}}.{\rm{k}}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{\sin {\rm{x}}}}{{{{\rm{k}}^2}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{{\rm{ak}}}}{{{{\rm{k}}^2}}} + \dfrac{{\cos \left( {1 + \cos {\rm{x}}} \right) + {{\sin }^2}{\rm{x}}}}{{\sin {\rm{x}}\left( {1 + \cos {\rm{x}}} \right)}}$
$ \Rightarrow \dfrac{{\rm{a}}}{{\rm{x}}} + \dfrac{1}{{\sin {\rm{x}}}}$
$ \Rightarrow \dfrac{{\rm{a}}}{{\rm{x}}} + \dfrac{1}{{{\rm{ax}}}}$
$ \Rightarrow \dfrac{1}{{\rm{k}}}\left[ {{\rm{a}} + \dfrac{1}{{\rm{a}}}} \right]$
So, from the above option only A option is correct.
Note:
In these questions try to substitute the values from the given equation in the asked expression. Use Trigonometric formulae to simplify the expression. It is recommended that you learn all the trigonometric formulae.
First of all we’ll find the values of a, b and c in terms of k and put the values in ${\rm{bc}} + \dfrac{1}{{{\rm{ck}}}} + \dfrac{{{\rm{ak}}}}{{1 + 6{\rm{k}}}}$. Then we’ll simplify further using trigonometric formulas to get the answer.
Complete step by step solution:
$\dfrac{{\sin {\rm{x}}}}{{\rm{a}}} = \dfrac{{\cos {\rm{x}}}}{{\rm{b}}} = \dfrac{{\tan {\rm{x}}}}{{\rm{c}}} = {\rm{k}}$ $\left[ {\dfrac{{\sin {\rm{x}}}}{{\rm{k}}} = {\rm{a}}} \right]$
${\rm{bc}} + \dfrac{1}{{{\rm{ck}}}} + \dfrac{{{\rm{ak}}}}{{1 + {\rm{bk}}}}$
We can write
$ \Rightarrow \dfrac{{\cos {\rm{x}}}}{{\rm{k}}} \times \dfrac{{\tan {\rm{x}}}}{{\rm{k}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{\cos {\rm{x}}}}{{\rm{k}}} \times \dfrac{{\sin {\rm{x}}}}{{\cos {\rm{x}}.{\rm{k}}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{\sin {\rm{x}}}}{{{{\rm{k}}^2}}} + \dfrac{1}{{\tan {\rm{x}}}} + \dfrac{{\sin {\rm{x}}}}{{1 + \cos {\rm{x}}}}$
$ \Rightarrow \dfrac{{{\rm{ak}}}}{{{{\rm{k}}^2}}} + \dfrac{{\cos \left( {1 + \cos {\rm{x}}} \right) + {{\sin }^2}{\rm{x}}}}{{\sin {\rm{x}}\left( {1 + \cos {\rm{x}}} \right)}}$
$ \Rightarrow \dfrac{{\rm{a}}}{{\rm{x}}} + \dfrac{1}{{\sin {\rm{x}}}}$
$ \Rightarrow \dfrac{{\rm{a}}}{{\rm{x}}} + \dfrac{1}{{{\rm{ax}}}}$
$ \Rightarrow \dfrac{1}{{\rm{k}}}\left[ {{\rm{a}} + \dfrac{1}{{\rm{a}}}} \right]$
So, from the above option only A option is correct.
Note:
In these questions try to substitute the values from the given equation in the asked expression. Use Trigonometric formulae to simplify the expression. It is recommended that you learn all the trigonometric formulae.
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