Question

Prove that: 1+ cos(theta)+ sin(theta) /1+cos(theta)- sin(theta)=1+ sin(theta) / cos(theta)

Answer 1

Dear student
$$\begin{gathered} \frac{1+\mathrm{cos\theta }+\mathrm{sin\theta }}{1+\mathrm{cos\theta }-\mathrm{sin\theta }} \\ =\frac{\left\{1+\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)\right\}\left\{1-\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)\right\}}{\left\{1+\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)\right\}\left\{1-\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)\right\}} \\ =\frac{1-\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)+\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)\left\{1-\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)\right\}}{1-{\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}^2} \\ =\frac{1-\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)+\mathrm{cos\theta }-\mathrm{cos\theta }\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)+\mathrm{sin\theta }-\mathrm{sin\theta }\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}{1-{\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}^2} \\ =\frac{1-\mathrm{cos\theta }+\mathrm{sin\theta }+\mathrm{cos\theta }-{\cos }^2\mathrm{\theta }+\mathrm{cos\theta }\mathrm{sin\theta }+\mathrm{sin\theta }-\mathrm{sin\theta }\mathrm{cos\theta }+{\sin }^2\mathrm{\theta }}{1-\left({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }-2\mathrm{sin\theta }\mathrm{cos\theta }\right)} \\ =\frac{1+2\mathrm{sin\theta }+{\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }}{1-1+2\mathrm{sin\theta }\mathrm{cos\theta }} \\ =\frac{1+2\mathrm{sin\theta }+{\sin }^2\mathrm{\theta }-\left(1-{\sin }^2\mathrm{\theta }\right)}{2\mathrm{sin\theta }\mathrm{cos\theta }} \\ =\frac{2\mathrm{sin\theta }+{\sin }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{2\mathrm{sin\theta }\mathrm{cos\theta }} \\ =\frac{2\mathrm{sin\theta }\left(1+\mathrm{sin\theta }\right)}{2\mathrm{sin\theta }\mathrm{cos\theta }} \\ =\frac{1+\mathrm{sin\theta }}{\mathrm{cos\theta }} \end{gathered}$$
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