MCQ
If $8\tan \text{x} = 15,$ then $\sin \text{x} - \cos \text{x}$ is equal to :
- A$\frac{8}{17}$
- B$\frac{17}{7}$
- C$\frac{1}{17}$
- ✓$\frac{7}{17}$
✓
Answer
Correct option: D.
$\frac{7}{17}$
We have,
$8\tan\text{x}=15$
$\Rightarrow\tan\text{x}=\frac{15}{8}$
In $\triangle\text{ABC,}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(15)^2+(8)^2$
$\Rightarrow\text{AC}^2=225+64$
$\Rightarrow\text{AC}^2=289$
$\Rightarrow\text{AC}=17$
$\therefore\sin\text{x}=\frac{15}{17} $ and $\cos\text{x}=\frac{8}{17}$
Now, $\sin\text{x}-\cos\text{x}=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
Hence the correct option is $(d)$
$8\tan\text{x}=15$
$\Rightarrow\tan\text{x}=\frac{15}{8}$
In $\triangle\text{ABC,}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(15)^2+(8)^2$
$\Rightarrow\text{AC}^2=225+64$
$\Rightarrow\text{AC}^2=289$
$\Rightarrow\text{AC}=17$
$\therefore\sin\text{x}=\frac{15}{17} $ and $\cos\text{x}=\frac{8}{17}$
Now, $\sin\text{x}-\cos\text{x}=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
Hence the correct option is $(d)$
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