Let −61 β1 and α2>β2, then α1+β2 equals

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Let -{{1}}{{6}}<<-{}{{12}} Suppose _{{1}}{{and}}_{{1}}, are th

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Let $- \frac{1}{6} < θ < - \frac{π}{12}$ Suppose $α_{1} and β_{1}$ , are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2} and β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

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Samuel Discussed

Let

- \frac{1}{6} < θ < - \frac{π}{12}

Suppose

α_{1} and β_{1}

, are the roots of the equation

x^{2} - 2 x sec θ + 1 = 0

and

α_{2} and β_{2}

are the roots of the equation

x^{2} + 2 x tan θ - 1 = 0

. If

α_{1} > β_{1}

and

α_{2} > β_{2}

, then

α_{1} + β_{2}

equals

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Question Text	Let $- \frac{1}{6} < θ < - \frac{π}{12}$ Suppose $α_{1} and β_{1}$ , are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2} and β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals
Answer Type	Video solution: 1
Upvotes	152