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Prove that (sec^4x - sec^2x) = tan^2 x+ tan^4x

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Solution: We have,

$\ LHS=sec^4x-tan^2x\ \ecause1+tan^2x=sec^2x$

Hence,

$\Rightarrow (sec^2x)^2-sec^2x \ \Rightarrow (1+tan^2x)^2-(1+tan^2x)\ \Rightarrow 1+tan^4x+2tan^2x-1-tan^2x\ \Rightarrow tan^4x+tan^2x=RHS$

Posted by

Deependra Verma

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