Question

Alpha+ beta =90 degree then prove sec square alpha + sec squre beta equal to sec squre alpha sec squre beta

Answer 1

Step-by-step explanation:

$given \: \alpha + \beta = {90}^{0} \\ \\ sec {}^{2} \alpha + sec {}^{2} \beta \\ \\ = \frac{1}{ {cos}^{2} \alpha } + \frac{1}{ {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {cos}^{2} \beta }{ {cos}^{2} \alpha {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {cos}^{2} (90 {}^{0} - \alpha ) }{ {cos}^{2} \alpha {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {sin}^{2} \alpha }{ {cos}^{2} \alpha {cos}^{2} \beta } = \frac{1}{ {cos}^{2} \alpha {cos}^{2} \beta } = {sec}^{2} \alpha {sec}^{2} \beta$

$formulae \\ \\ cosx \: secx \: = \: 1 \\ \\ cos ( {90}^{0} - x) = sinx \\ \\ {cos}^{2} x + {sin}^{2} x = 1$

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Answer 2

Step-by-step explanation:

$\begin{lgathered}given \: \alpha + \beta = {90}^{0} \\ \\ sec {}^{2} \alpha + sec {}^{2} \beta \\ \\ = \frac{1}{ {cos}^{2} \alpha } + \frac{1}{ {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {cos}^{2} \beta }{ {cos}^{2} \alpha {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {cos}^{2} (90 {}^{0} - \alpha ) }{ {cos}^{2} \alpha {cos}^{2} \beta } \\ \\ = \frac{ {cos}^{2} \alpha + {sin}^{2} \alpha }{ {cos}^{2} \alpha {cos}^{2} \beta } = \frac{1}{ {cos}^{2} \alpha {cos}^{2} \beta } = {sec}^{2} \alpha {sec}^{2} \beta \\ \\\end{lgathered}$