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, \[=\frac{{{\left( \frac{12}{13} \right)}^{2}}-{{\left( \frac{5}{13} \right)}^{2}}}{2\left( \frac{12}{13} \right)\left( \frac{5}{13} \right)}\times \frac{1}{{{\left( \frac{12}{5} \right)}^{2}}}\] \[=\frac{\frac{144-25}{169}}{\frac{120}{169}}\times \frac{25}{144}\] \[=\frac{119}{120}\times \frac{25}{144}=\frac{595}{3456}\]
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