Question

$\frac{\mathrm{sin\theta }}{(1-\mathrm{cot\theta })}+\frac{\mathrm{cos\theta }}{(1-\mathrm{tan\theta })}=?$
(a) (cos θ + sin θ)
(b) (cos θ − sin θ)
(c) 0
(d) 2 tan θ

Answer 1

(a) (cos θ + sin θ)

$$\begin{gathered} \frac{\text{sin}\theta }{(1-c\text{ot}\theta )}+\frac{\text{cos}\theta }{(1-\text{tan}\theta )} \\ \text{=}\frac{\text{sin}\theta }{1-\frac{\text{cos}\theta }{\text{sin}\theta }}+\frac{\text{cos}\theta }{1-\frac{\text{sin}\theta }{\text{cos}\theta }} \\ =\left(\frac{\text{sin}\theta }{\frac{\text{sin}\theta -c\text{os}\theta }{\text{sin}\theta }}\right)+\left(\frac{\text{cos}\theta }{\frac{\text{cos}\theta -s\text{in}\theta }{\text{cos}\theta }}\right) \\ \text{=}\left(\frac{{\text{sin}}^2\theta }{\text{sin}\theta -c\text{os}\theta }\right)+\left(\frac{{\text{cos}}^2\theta }{\text{cos}\theta -s\text{in}\theta }\right) \\ \text{=}\left(\frac{{\text{cos}}^2\theta }{\text{cos}\theta -s\text{in}\theta }\right)-\left(\frac{{\text{sin}}^2\theta }{\text{cos}\theta -s\text{in}\theta }\right) \\ \text{=}\left(\frac{{\text{cos}}^2\theta -s{\text{in}}^2\theta }{\text{cos}\theta -s\text{in}\theta }\right) \\ \text{=}\left\{\frac{\left(c\text{os}\theta \text{+sin}\theta \right)(c\text{os}\theta -s\text{in}\theta )}{(c\text{os}\theta -s\text{in}\theta )}\right\} \\ \text{=cos}\theta +s\text{in}\theta \end{gathered}$$