Question

answer the question ​

Answer 1

HIII.. .. .. .. ..

THERE IS YOUR ANSWER.... .

$since \: \sin( a ) \: and \: \cos( a ) \: is \: the \: \\ rootsof \: the \: equation \: \\ 4 {x}^{2} - 3x + \alpha = 0 \\ \\ than \: \sin(a) + \cos(a) = \frac{ - ( - 3)}{4} \\ \\ divide \: the \: equation \: by \: cos(a) \\ than \: the \: equation \: become \\ \\ \tan(a) + 1 = \frac{3}{4} ...........eq(2) \\ \\ \tan(a) = \frac{ - 1}{4} \\ \\ from \: first \: equation \\ \\ \sin(a) + \cos(a) = \frac{3}{4} \\ \\ \\ now \: we \: can \: use \: the \: question..... \\ \\ \sin(a ) + \cos(a) + \tan(a) + \cot(a) + \sec(a) + \csc(a) = 7 \\ \\ after \: using \: all \: the \: above \: eq. \\ \\ \frac{3}{4} + \frac{ - 1}{4} + ( - 4) + \frac{ \sin(a) + cos(a) }{ \sin(a) \cos(a) } = 7 \\ \\ \sin(a) \cos(a) = \frac{ \alpha }{4} \\ \frac{3}{4} + \frac{ - 1}{4} - 4 + \frac{ \frac{3}{4} }{ \frac{ \alpha }{4} } = 7 \\ \\ \frac{2}{4} - 4 + \frac{3}{ \alpha } = 7 \\ \\ multiply \: the \: whole \: by \: 4 \alpha \\ \\ than \: we \: get \\ 2 \alpha - 16 \alpha + 12 = 28 \alpha \\ \\ - 14 \alpha + 12 = 28 \alpha \\ \\ 12 = 42 \alpha \\ \alpha = \frac{2}{7} \\ \\ now \: the \: value \: of \: | \frac{25 \alpha }{4} | \\ \frac{50}{28}$

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