MCQ
If $8 \tan x = 15,$ then $\sin x - \cos x$ is equal to
- A$\frac{17}{7}$
- B$\frac{8}{17}$
- ✓$\frac{7}{17}$
- D$\frac{1}{17}$
✓
Answer
Correct option: C.
$\frac{7}{17}$
$8 \tan x =15$
$\Rightarrow \tan x=\frac{15}{8}=\frac{\text { Perpendicular }}{\text { Base }}$
By Pythagoras Theorem,
$(\text { Hyp. })^2=(\text { Base })^2+(\text { Perp. })^2$
$=(8)^2+(15)^2$
$=64+225=289=(17)^2$
$\therefore \text { Hyp. }=17 \text { units }$
$\therefore \sin x=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{15}{17}$
$\cos x=\frac{\text { BBase }}{\text { Hypotenuse }}=\frac{8}{17}$
$\sin x-\cos x=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
$\Rightarrow \tan x=\frac{15}{8}=\frac{\text { Perpendicular }}{\text { Base }}$
By Pythagoras Theorem,
$(\text { Hyp. })^2=(\text { Base })^2+(\text { Perp. })^2$
$=(8)^2+(15)^2$
$=64+225=289=(17)^2$
$\therefore \text { Hyp. }=17 \text { units }$
$\therefore \sin x=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{15}{17}$
$\cos x=\frac{\text { BBase }}{\text { Hypotenuse }}=\frac{8}{17}$
$\sin x-\cos x=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
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