Question

BRAINLEST FOR BEST ​

Answer 1

This is the answer

Hope i will get the brainliest

Answer 2

$we \: know \: that \\ \sin( \alpha ) = \cos( 90 - \beta ) \\ here \\ \sin(43) = \sin(90 - 47) \\ = \cos(47) \\ so \: in \: the \: first \\ {( \frac{3 \sin(43) }{ \cos(47) }) }^{2} \\ = { (\frac{3 \cos(47) }{ \cos(47) } )}^{2} = 9\\ then \: there \: has \: a \: formula \\ of \: cosec \alpha \: and \: sec \alpha \\ cosec \alpha = \sec(90 - \beta ) \\ so \: cosec53 \\ = cosec(90 - 47) \\ = sec47 \\ = \frac{1}{ \cos(37) } \\ then \\ \cos(37) \times \csc(53) \\ = \cos(37) \times \frac{1}{ \cos(37) } \\ = 1 \\ we \: know \: that \: the \: value \: of \\ tan45 = 1 \\ \tan(5) = \tan(90 - 85) \\ = \cot(85) = \frac{1}{ \tan(85) } \\ now \\ \tan(25) = \tan(90 - 65) \\ = \cot(65) = \frac{1}{ \tan(65) } \\ then \: in \: the \: question \\ \tan5 \tan25\tan45\tan65 \tan85 \\ = \tan(5) \tan(85) \tan(25) \tan(65) \times 1 \\ because \: tan45 = 1 \\ = \frac{1}{ \tan(85) } \tan(85) \frac{1}{ \tan(65) } \tan(65) \\ = 1 \\ so \: it \: stands \: for \\ 9- \frac{1}{1} \\ = 9- 1 = 8$

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