Question

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Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

✴$\red{\:1}$.

$\large{\underline{\underline{\mathfrak{\pink{\sf{Given\:Here:-}}}}}}$.

✴$\red{\frac{\:sec^4\theta\:-\:tan^2\theta}{\:sec^2\theta+\:tan^2\theta}}$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Explanation:-}}}}}}$.

$\implies\frac{(\:sec^4\theta\:-\:tan^4\theta)}{(\:sec^2\theta\:+\:tan^2\theta)}$.

$\implies\frac{\:(sec^2\theta)^2\:-\:(tan^2\theta)^2}{\:sec^2\theta\:+\:tan^2\theta}$.

➡We know that ,

✴$\red{\:(a^2-b^2)\:=\:(a+b)(a-b)}$.

➡Using this identity .

$\implies\frac{(\:sec^2\theta\:+\:tan^2\theta)(\:sec^2\theta\:-\:tan^2\theta)}{\:sec^2\theta\:+\:tan^2\theta}$.

$\implies\:(sec^2\theta\:-\:tan^2\theta)$.

➡We Know,

✴$\red{(\:sec^2\theta\:-\:tan^2\theta)\:=\:1}$.

$\implies\:1$.

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