MCQ
If $\text{cosec x}+\cot\text{x}=\frac{11}{2},$ then $\tan\text{x}=$
- A$\frac{21}{22}$
- B$\frac{15}{16}$
- ✓$\frac{44}{117}$
- D$\frac{117}{43}$
✓
Answer
Correct option: C.
$\frac{44}{117}$
We have:
$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$
$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec}\text{ x}-\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$
$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$
subtracting $(2)$ from $(1):$
$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$
$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$
$\Rightarrow2\cot\text{x}=\frac{117}{22}$
$\Rightarrow\cot\text{x}=\frac{117}{44}$
$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$
$\Rightarrow\tan\text{x}=\frac{44}{117}$
$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$
$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec}\text{ x}-\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$
$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$
subtracting $(2)$ from $(1):$
$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$
$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$
$\Rightarrow2\cot\text{x}=\frac{117}{22}$
$\Rightarrow\cot\text{x}=\frac{117}{44}$
$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$
$\Rightarrow\tan\text{x}=\frac{44}{117}$
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