Question
If $\tan\theta=\frac{\text{a}}{\text{b}} ,$ find the value of $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.$
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Answer
$\tan\theta=\frac{\text{a}}{\text{b}}$ find $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$
Divide equation (i) with $\cos\theta,$ we get
$\Rightarrow\frac{\frac{\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\frac{\sin\theta}{\cos\theta}}{ 1-\frac{\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\tan\theta}{1-\tan\theta}$
$=\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}} $
$=\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
Divide equation (i) with $\cos\theta,$ we get
$\Rightarrow\frac{\frac{\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\frac{\sin\theta}{\cos\theta}}{ 1-\frac{\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\tan\theta}{1-\tan\theta}$
$=\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}} $
$=\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
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