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Question

If $$\cot \theta =\dfrac {7}{8}$$, evaluate:
$$\dfrac {(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$$.


Solution

$$\begin{array}{l} \cot  \theta =\dfrac { 7 }{ 8 }  \\ \dfrac { { \left( { 1+sin\theta  } \right) \left( { 1-sin\theta  } \right)  } }{ { \left( { 1+\cos  \theta  } \right) \left( { 1-\cos  \theta  } \right)  } }  \\ =\dfrac { { { 1^{ 2 } }-{ sin^{ 2 } }\theta  } }{ { { 1^{ 2 } }-{ { \cos   }^{ 2 } }\theta  } } =\dfrac { { { { \cos   }^{ 2 } }\theta  } }{ { { { \sin   }^{ 2 } }\theta  } }  \\ ={ \cot ^{ 2 }  }\theta ={ \left( { \dfrac { 7 }{ 8 }  } \right) ^{ 2 } }=\dfrac { { 49 } }{ { 64 } }  \end{array}$$

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