question_answer

If \[\sin \theta =\frac{12}{13},0{}^\circ <\theta <90{}^\circ ,\], find the value of: \[\frac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta .\cos \theta }\times \frac{1}{{{\tan }^{2}}\theta }\]

Answer:

Given, sin \[\theta =\frac{12}{13}\]

\[\Rightarrow \frac{P}{H}=\frac{12}{13}\]

Let, \[P=12K,\text{ }H=13K\]

\[{{P}^{2}}+{{B}^{2}}={{H}^{2}}\]         [Pythagoras theorem]

\[{{(12K)}^{2}}+{{B}^{2}}={{(13K)}^{2}}\]

\[144{{K}^{2}}+{{B}^{2}}=169{{K}^{2}}\]

\[{{B}^{2}}=169{{K}^{2}}-144{{K}^{2}}\]

\[=25{{K}^{2}}\]

\[B=5K\]

\[\therefore \cos \theta =\frac{B}{H}=\frac{5K}{13K}=\frac{5}{13}\]

\[\tan \theta =\frac{P}{B}=\frac{12K}{5K}=\frac{12}{5}\]

Now, \[\frac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta .\cos \theta }\times \frac{1}{{{\tan }^{2}}\theta }\]

On solving,

\[=\frac{{{\left( \frac{12}{13} \right)}^{2}}-{{\left( \frac{5}{13} \right)}^{2}}}{2\left( \frac{12}{13} \right)\left( \frac{5}{13} \right)}\times \frac{1}{{{\left( \frac{12}{5} \right)}^{2}}}\]

\[=\frac{\frac{144-25}{169}}{\frac{120}{169}}\times \frac{25}{144}\]

\[=\frac{119}{120}\times \frac{25}{144}=\frac{595}{3456}\]