MCQ
If $\tan\theta=\frac{\text{a}}{\text{b}}$ then $\frac{(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)}=?$
  • A
    $\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$
  • B
    $\frac{\text{a}-\text{b}}{\text{a}+\text{b}}$
  • $\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
  • D
    $\frac{\text{b}-\text{a}}{\text{b}+\text{a}}$

Answer

Correct option: C.
$\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
Given, $\tan\theta=\frac{\text{a}}{\text{b}}$
Now, $\frac{(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)}=\frac{\frac{\cos\theta}{\cos\theta}+\frac{sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}$
$=\frac{1+\tan\theta}{1-\tan\theta}$
$=\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}}$
$=\frac{\frac{\text{b}+\text{a}}{\text{b}}}{\frac{\text{b}-\text{a}}{\text{b}}}$
$=\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$

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