Question

If $\cos \theta =\frac{3}{5}$, show that $\frac{\left(\sin \theta -\cot \theta \right)}{2\tan \theta }=\frac{3}{160}$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\left(\sin \theta -\cot \theta \right)}{2\tan \theta } \\ =\frac{\left(\sin \theta -{\displaystyle \frac{\cos \theta }{\sin \theta }}\right)}{2\left({\displaystyle \frac{\sin \theta }{\cos \theta }}\right)} \\ =\frac{\left({\displaystyle \frac{{\sin }^2\theta -\cos \theta }{\sin \theta }}\right)}{\left({\displaystyle \frac{2\sin \theta }{\cos \theta }}\right)} \\ =\frac{\cos \theta \left({\sin }^2\theta -\cos \theta \right)}{2{\sin }^2\theta } \\ =\frac{\cos \theta \left(1-{\cos }^2\theta -\cos \theta \right)}{2\left(1-{\cos }^2\theta \right)} \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle \frac{3}{5}\left[1-{\left(\frac{3}{5}\right)}^2-\frac{3}{5}\right]}}{2\left[1-{\left({\displaystyle \frac{3}{5}}\right)}^2\right]} \\ =\frac{{\displaystyle \frac{3}{5}\left(\frac{1}{1}-\frac{9}{25}-\frac{3}{5}\right)}}{2\left(1-{\displaystyle \frac{9}{25}}\right)} \\ =\frac{{\displaystyle \frac{3}{5}\left(\frac{25-9-15}{25}\right)}}{2\left({\displaystyle \frac{25-9}{25}}\right)} \\ =\frac{{\displaystyle \frac{3}{5}\left(\frac{1}{25}\right)}}{2\left({\displaystyle \frac{16}{25}}\right)} \end{gathered}$$
$$\begin{gathered} =\frac{3}{5\times 2\times 16} \\ =\frac{3}{160} \\ =\mathrm{RHS} \end{gathered}$$