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Question

Prove that
$$\dfrac { \tan ^{ 2 }{ \theta  }  }{ { \left( \sec { \theta  } -1 \right)  }^{ 2 } } =\dfrac { 1+\cos { \theta  }  }{ 1-\cos { \theta  }  } $$


Solution

L.H.S

$$\Rightarrow \dfrac{{{\tan }^{2}}\theta }{{{\left( \sec \theta -1 \right)}^{2}}}$$

$$ \Rightarrow \dfrac{\left( {{\sec }^{2}}\theta -1 \right)}{{{\left( \sec \theta -1 \right)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta  \right] $$

$$ \Rightarrow \dfrac{\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)}{{{\left( \sec \theta -1 \right)}^{2}}} $$

$$ \Rightarrow \dfrac{\left( \sec \theta +1 \right)}{\left( \sec \theta -1 \right)} $$

$$ \Rightarrow \dfrac{\left( \dfrac{1}{\cos \theta }+1 \right)}{\left( \dfrac{1}{\cos \theta }-1 \right)} $$

$$ \Rightarrow \dfrac{\left( 1+\cos \theta  \right)}{\left( 1-\cos \theta  \right)}\, $$

 

R.H.S

Hence, proved.


Mathematics

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