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Question

Prove that
$$2\sec^{2}\theta -\sec^{4}\theta -2\csc^{2}\theta +\csc^{4}\theta =\cot^{4}\theta -\tan^{4}\theta$$


Solution

$$\begin{array}{l} 2{ \sec ^{ 2 }  }\theta -{ \sec ^{ 4 }  }\theta -2{ \csc ^{ 2 }  }\theta +{ \csc^{ 4 } }\theta ={ \cot ^{ 4 }  }\theta -{ \tan ^{ 4 }  }\theta  \\ 2\left( { 1+{ { \tan   }^{ 2 } }\theta  } \right) -{ \left( { 1+{ { \tan   }^{ 2 } }\theta  } \right) ^{ 2 } }-2\left( { 1+{ { \cot   }^{ 2 } }\theta  } \right) +{ \left( { 1+{ { \cot   }^{ 2 } }\theta  } \right) ^{ 2 } } \\ =2+2{ \tan ^{ 2 }  }\theta -\left( { 1+{ { \tan   }^{ 4 } }\theta +2{ { \tan   }^{ 2 } }\theta  } \right) -\left( { 2+2{ { \cot   }^{ 2 } }\theta  } \right) +1+{ \cot^{ 4 } }\theta +2{ \cot ^{ 2 }  }\theta  \\ =2+2{ \tan ^{ 2 }  }\theta -1-{ \tan ^{ 4 }  }\theta -2\tan^2{\theta}-2-2{ \cot ^{ 2 }  }\theta +1+{ \cot ^{ 4 }  }\theta +2{ \cot ^{ 2 }  }\theta  \\ ={ \cot ^{ 4 }  }\theta -{ \tan ^{ 4 }  }\theta \,\\ \, \end{array}$$

R.H.S
Hence, proved.

Maths

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