Question

3cosΦ=2
2sec^2Φ+2tan^2Φ-7=?

Answer 1

Answer:

given

$3 \cos( \theta) = 2 \\ \cos( \theta) = \frac{3}{2}$

$\sec( \theta) = \frac{2}{3}$

we know that

$\tan( \theta) = \sqrt{1 - { \sec( \theta) }^{2} }$

$\tan( \theta ) = \sqrt{1 - { (\frac{2}{3} )}^{2} } \\ \tan( \theta) = \sqrt{ \frac{9 - 4}{9} } \\ \tan( \theta) = \sqrt{ \frac{5}{9} } \\ \tan( \theta) = \frac{ \sqrt{5} }{3}$

${\mathrm{\underline{\underline \pink {to \: \: find:}}}}$

$2 { \sec( \theta) }^{2} + 2 { \tan( \theta) }^{2} - 7$

$= 2 {( \frac{2}{3}) }^{2} + 2 {( \frac{ \sqrt{5} }{3} )}^{2} \\ \\ = 2( \frac{4}{9} ) + 2( \frac{5}{9}) \\ \\ = \frac{8}{9} + \frac{10}{9} \\ \\ = \frac{18}{9} \\ = 2$

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