Question
If $\tan\theta=\frac{4}{5},$ find the value of $\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}.$
✓
Answer
We have, $\tan\theta=\frac{4}{5}$ In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(4)^2+(5)^2$ $\Rightarrow\text{AC}^2=16+25$ $\Rightarrow\text{AC}^2=41$ $\Rightarrow\text{AC}=\sqrt{41}$ $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{\sqrt{14}}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4}{\sqrt{41}}$ Now, $\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}$ $=\frac{\frac{5-4}{\sqrt{41}}}{\frac{5+4}{\sqrt{41}}}$ $=\frac{1}{9}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(4)^2+(5)^2$ $\Rightarrow\text{AC}^2=16+25$ $\Rightarrow\text{AC}^2=41$ $\Rightarrow\text{AC}=\sqrt{41}$ $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{\sqrt{14}}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4}{\sqrt{41}}$ Now, $\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}$ $=\frac{\frac{5-4}{\sqrt{41}}}{\frac{5+4}{\sqrt{41}}}$ $=\frac{1}{9}$
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