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

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