Question
If $\sec\theta=\frac{13}{5},$ show that $\frac{\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=3.$
✓
Answer
We have,
$\sec\theta=\frac{13}{5}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(13)^2=\text{AB}^2+(5^2)$
$\Rightarrow\text{AB}^2=169-25$
$\Rightarrow\text{AB}^2=144$
$\Rightarrow\text{AB}=12$
$\therefore\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$ and $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
Now, $\frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=\frac{2\times\frac{12}{13}-3\times\frac{5}{13}}{4\times\frac{12}{13}-9\times\frac{5}{13}}$
$=\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
$=\frac{9}{3}$
$=3$
$\sec\theta=\frac{13}{5}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(13)^2=\text{AB}^2+(5^2)$
$\Rightarrow\text{AB}^2=169-25$
$\Rightarrow\text{AB}^2=144$
$\Rightarrow\text{AB}=12$
$\therefore\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$ and $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
Now, $\frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=\frac{2\times\frac{12}{13}-3\times\frac{5}{13}}{4\times\frac{12}{13}-9\times\frac{5}{13}}$
$=\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
$=\frac{9}{3}$
$=3$
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