Question

If $\sec \theta =\frac{5}{4}$, show that $\frac{\left(\sin \theta -2\cos \theta \right)}{\left(\tan \theta -\cot \theta \right)}=\frac{12}{7}$.

Answer 1

We have,

$$\begin{gathered} \sec \theta =\frac{5}{4} \\ \Rightarrow \frac{1}{\cos \theta }=\frac{5}{4} \\ \Rightarrow \cos \theta =\frac{4}{5} \end{gathered}$$

Also,

$$\begin{gathered} {\sin }^2\theta =1-{\cos }^2\theta \\ =1-{\left(\frac{4}{5}\right)}^2 \\ =1-\frac{16}{25} \\ =\frac{9}{25} \\ \Rightarrow \sin \theta =\frac{3}{5} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{LHS}=\frac{\left(\sin \theta -2\cos \theta \right)}{\left(\tan \theta -\cot \theta \right)} \\ =\frac{\left(\sin \theta -2\cos \theta \right)}{\left({\displaystyle \frac{\sin \theta }{\cos \theta }}-{\displaystyle \frac{\cos \theta }{\sin \theta }}\right)} \\ =\frac{\left(\sin \theta -2\cos \theta \right)}{\left({\displaystyle \frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta }}\right)} \\ =\frac{\sin \theta \cos \theta \left(\sin \theta -2\cos \theta \right)}{\left({\displaystyle {\sin }^2\theta -{\cos }^2\theta }\right)} \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle \frac{3}{5}}\times {\displaystyle \frac{4}{5}}\left({\displaystyle \frac{3}{5}}-2\times {\displaystyle \frac{4}{5}}\right)}{{\left({\displaystyle \frac{3}{5}}\right)}^2-{\left({\displaystyle \frac{4}{5}}\right)}^2} \\ =\frac{{\displaystyle \frac{12}{25}\left(\frac{3}{5}-\frac{8}{5}\right)}}{\left({\displaystyle \frac{9}{25}}-{\displaystyle \frac{16}{25}}\right)} \\ =\frac{{\displaystyle \frac{12}{25}\times \left(\frac{-5}{5}\right)}}{\left({\displaystyle \frac{-7}{25}}\right)} \\ =\frac{{\displaystyle 12}}{7} \\ =\mathrm{RHS} \end{gathered}$$