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# If $16\cot x = 12$ find the value of $\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$A.7B.$\dfrac{7}{3}$C.$\dfrac{7}{2}$D.None of these

Last updated date: 13th Jun 2024
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Hint: For the given question we have to divide denominator and numerator with $\sin x$. It is known that $\dfrac{{\cos x}}{{\sin x}} = \cot x$. Then it is given that $16\cot x = 12$, we will substitute the value of $\cot x$ and after simplifying we the value of the given question.

Given $16\cot x = 12$simplifying, we get,
$\Rightarrow \cot x = \dfrac{3}{4}$ ……………. $\left( i \right)$
Now according to question
$\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$
Dividing denominator and numerator with $\sin x$,we get,
$\Rightarrow \left[ {\dfrac{{1 + \dfrac{{\cos x}}{{\sin x}}}}{{1 - \dfrac{{\cos x}}{{\sin x}}}}} \right]$
Substituting $\dfrac{{\cos x}}{{\sin x}} = \cot x$, we get,
$\Rightarrow \left[ {\dfrac{{1 + \cot x}}{{1 - \cot x}}} \right]$
Now, substituting $\left( i \right)$and on simplifying we get,
$\Rightarrow \left[ {\dfrac{{1 + \dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}} \right]$
$= 7$
$\therefore \left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right] = 7$
$\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$
Dividing denominator and numerator with$\cos x$we get
$\Rightarrow \left[ {\dfrac{{\tan x + 1}}{{\tan x - 1}}} \right]$
We know that $\cot x = \dfrac{3}{4}$and $\tan x = \dfrac{1}{{\cot x}}$, we get,
$\therefore \tan x = \dfrac{4}{3}$………….. (ii)
$\Rightarrow \left[ {\dfrac{{\dfrac{4}{3} + 1}}{{\dfrac{4}{3} - 1}}} \right] = 7$
$\therefore \left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right] = 7$