Question

if sec A = root 2 find .3 cos square A +5 tan square A divided by 4 tan square A - sin square A.​

Answer 1

Step-by-step explanation:

$\sec(a) = \sqrt{2} \\ \frac{3 \cos {}^{2} (a) + 5 \tan {}^{2} (a) }{4 \tan {}^{2} (a) - \sin {}^{2} (a) } \\ sec \: a = \frac{hypotenuse}{base} = \frac{ \sqrt{2} }{1} \\ so \: hypotenuse \: \sqrt{2} and \: base = 1 \\ perpendicular = 1(pythagoras) \\ cos \: a = \frac{base}{hypotenuse} = \frac{1}{ \sqrt{2} } \\ tan \: a = \frac{perpendicular}{base} = \frac{1}{1} = 1 \\ sin \: a = \frac{perpendicular}{hypotenuse} = \frac{1 } { \sqrt{2} } \\ put \: the \: values \\ = \frac{3 \times \frac{1}{2} + 5 \times 1}{4 \times 1 - \frac{1}{2} } \\ = \frac{ \frac{3}{2} + 5}{4 - \frac{1}{2} } \\ = \frac{ \frac{13}{2} }{ \frac{7}{2} } \\ = \frac{13}{7}$

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