Question

prove that (sinA+secA)² + (cosA+cosecA)² = 1-secA×cosA
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Answer 1

$( {sinA + secA})^{2} + (cosA + cosecA) ^{2} \\ \\ = ( {sinA + \frac{1}{cosA} })^{2} + ( {cosA + \frac{1}{sinA} })^{2} \\ \\ = {sin}^{2} A + \frac{1}{ {cos}^{2} A} + 2 \frac{sinA}{cosA} + {cos}^{2} A + \frac{1}{ {sin}^{2} A} + 2 \frac{cosA}{sinA} \\ \\ = 1 + ( \frac{1}{ {sin}^{2}A } + \frac{1}{ {cos}^{2} A} ) + 2( \frac{sinA}{cosA} + \frac{cosA}{sinA} ) \\ \\ = 1 + \frac{1}{ {sin}^{2} A {cos}^{2}A } + 2 (\frac{1}{sinAcosA} ) \\ \\ = 1 + {sec}^{2} A {cosec}^{2} A + 2secAcosecA \\ \\ = ( {1 + secAcosecA})^{2}$