Question

If $\tan \mathrm{\theta }=\frac{1}{2}$ then evaluate $\left(\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}\right)$

Answer 1

$$\begin{gathered} \mathrm{Given}:\tan \theta =\frac{1}{2} \\ \mathrm{Since},\tan \theta =\frac{P}{B} \\ \Rightarrow P=1\mathrm{and}B=2 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+2^2=H^2 \\ \Rightarrow H^2=1+4 \\ \Rightarrow H^2=5 \\ \Rightarrow H=\sqrt{5} \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{\sqrt{5}} \\ \cos \theta =\frac{B}{H}=\frac{2}{\sqrt{5}} \\ \mathrm{Now}, \\ \left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\left(\frac{{\displaystyle \frac{2}{\sqrt{5}}}}{{\displaystyle \frac{1}{\sqrt{5}}}}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{1+{\displaystyle \frac{2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{{\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\right) \\ =\left(2+\left(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}\right)\right) \\ =\left(2+\left(\frac{\sqrt{5}-2}{5-4}\right)\right) \\ =\left(2+\sqrt{5}-2\right) \\ =\sqrt{5} \\ \mathrm{Hence},\left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\sqrt{5}. \end{gathered}$$