Question

prove that.
${sec}^{2} \alpha = 1 + { \tan \alpha }^{2}$


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Answer 1

Answer:

$1 + \frac{ {p}^{2} }{ {b}^{2} } \\ \frac{ {b}^{2} + {p}^{2} }{ {b}^{2} } \\ \frac{ {h}^{2} }{ {b}^{2} } = { \sec }^{2}$

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Answer 2

Answer:

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Step-by-step explanation:

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