10th Class Mathematics Solved Paper - Mathematics-2015 Term-I

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


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