MCQ
If $16\cot\times=12,$ then $\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $ equals :
- ✓$\frac{1}{7}$
- B$\frac{3}{7}$
- C$\frac{2}{7}$
- D$0$
✓
Answer
Correct option: A.
$\frac{1}{7}$
We are given $16\cot\text{x}=12.$We are asked to find the following
$\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $
We know that: $\cot\text{x}=\frac{\text{Base}}{\text{Perpendicular}}$
$\Rightarrow{\text{Base}}=3$
$\Rightarrow\text{Perpendicular}=4$
$\Rightarrow \text{Hypotenuse}=\sqrt{(\text{Perpendicular)}^2+\text{(Base)}^2}$
$\Rightarrow \text{Hypotenuse}=\sqrt{16+9}$
$\Rightarrow \text{Hypotenuse}=5$
Now we have
$16\cot\text{x}=12$
$\cot\text{x}=\frac{12}{16}$
$\cot\text{x}=\frac{3}{4}$
We know $\sin\text{x}=\frac{ \text{Perpendicular}}{ \text{Hypotenuse}}$ and $\cos\text{x}=\frac{ \text{Base}}{ \text{Hypotenuse}}$
Now we find
$\frac{\sin\text{x-\cos}\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}$
$=\frac{\frac{1}{5}}{\frac{7}{5}}$
$=\frac{1}{7}$
Hence the correct option is $(a)$
$\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $
We know that: $\cot\text{x}=\frac{\text{Base}}{\text{Perpendicular}}$
$\Rightarrow{\text{Base}}=3$
$\Rightarrow\text{Perpendicular}=4$
$\Rightarrow \text{Hypotenuse}=\sqrt{(\text{Perpendicular)}^2+\text{(Base)}^2}$
$\Rightarrow \text{Hypotenuse}=\sqrt{16+9}$
$\Rightarrow \text{Hypotenuse}=5$
Now we have
$16\cot\text{x}=12$
$\cot\text{x}=\frac{12}{16}$
$\cot\text{x}=\frac{3}{4}$
We know $\sin\text{x}=\frac{ \text{Perpendicular}}{ \text{Hypotenuse}}$ and $\cos\text{x}=\frac{ \text{Base}}{ \text{Hypotenuse}}$
Now we find
$\frac{\sin\text{x-\cos}\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}$
$=\frac{\frac{1}{5}}{\frac{7}{5}}$
$=\frac{1}{7}$
Hence the correct option is $(a)$
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