If $\cos e c x+\cot x=\frac{11}{2}$, then $\tan x=$
(a) $\frac{21}{22}$
(b) $\frac{15}{16}$
(c) $\frac{44}{117}$
(d) $\frac{117}{43}$
(c) $\frac{44}{117}$
We have:
$\operatorname{cosec} x+\cot x=\frac{11}{2}$ ....(i)
$\Rightarrow \frac{1}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$
$\Rightarrow \frac{\operatorname{cosec}^{2} x-\cot ^{2} x}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$
$\Rightarrow \frac{(\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x)}{(\operatorname{cosec} x+\cot x)}=\frac{2}{11}$
$\therefore \operatorname{cosec} x-\cot x=\frac{2}{11}$ ...(ii)
Subtracting (2) from (1):
$2 \cot x=\frac{11}{2}-\frac{2}{11}$
$\Rightarrow 2 \cot x=\frac{121-4}{22}$
$\Rightarrow 2 \cot x=\frac{117}{22}$
$\Rightarrow \cot x=\frac{117}{44}$
$\Rightarrow \frac{1}{\tan x}=\frac{117}{44}$
$\Rightarrow \tan x=\frac{44}{117}$
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