Question

prove that
$\sqrt{1 + \sin(a) } \div \sqrt{1 - \sin(a) } + \sqrt{1 - \sin(a) } \div \sqrt{1 + \sin(a) } = 2 \sec(a)$
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Answer 1

LHS=$\sqrt{1 + \sin(a) } \div \sqrt{1 - \sin(a) } + \sqrt{1 - \sin(a) } \div \sqrt{1 + \sin(a) }$

=$\frac{\left (\sqrt{1 + \sin(a)\right )^{2}+\left( \sqrt{1 - \sin(a)\right)^{2}}{\sqrt{1 + \sin(a)\times\sqrt{1-\sin(a)}}$

=$\frac{1+\sin(a)+1-\sin(a)}{\sqrt{1^{2}- \sin^{2}(a)}}$

=$\frac {2}{\sqrt {1-\sin^{2}(a)}}$

=$\frac {2}{\sqrt {cos^{2}(a)}}$

=$\frac{2}{cos(a)}$

= $2\sec(a)$

= RHS

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