Question

If tan θ = $\frac{a}{b},$ then $\frac{(\mathrm{cos\theta }+\mathrm{sin\theta })}{(\mathrm{cos\theta }-\mathrm{sin\theta })}=?$
(a) $\frac{a+b}{a-b}$
(b) $\frac{a-b}{a+b}$
(c) $\frac{b+a}{b-a}$
(d) $\frac{b-a}{b+a}$

Answer 1

(c) $\frac{b+a}{b-a}$
$$\begin{gathered} \text{Given: tan}\theta =\frac{a}{b} \\ \text{Now,}\frac{(\text{cos}\theta +\text{sin}\theta )}{(\text{cos}\theta -\text{sin}\theta )} \\ \text{=}\frac{(1+\text{tan}\theta )}{(1-\text{tan}\theta )}\left[\text{Dividing the numerator and denominator by cos}\theta \right] \\ \text{=}\frac{(1+\frac{a}{b})}{(1-\frac{a}{b})} \\ \text{=}\frac{\left(\frac{b+a}{b}\right)}{\left(\frac{b-a}{b}\right)} \\ \text{=}\frac{(b+a)}{(b-a)} \end{gathered}$$