Question

Help me guys...Pls....​

Answer 1

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$\frac{2 \tan(30) }{1 - { \tan(30) }^{2} } + 3 \sin(54) . \sec(54)$

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taking first term

it is in the form of

$\frac{2 \tan( \beta ) }{1 - { \tan( \beta ) }^{2} } = \tan2( \beta )$

now,

$\tan2(30) + 3 \sin(54) + \sec(36)$

$\tan(60) + 3 \sin(54) . \sec(36)$

we know that

$\tan(60) = \frac{ \sqrt{3} }{2} \\ \\ \sin(54) = \frac{ \sqrt{5} + 1}{4} \\ \\ \sec(36) = \frac{4}{ \sqrt{5} + 1}$

now, substitute in the question

$= \frac{ \sqrt{3} }{2} + 3( \frac{ \sqrt{5} + 1 }{4} )( \frac{4}{ \sqrt{5} + 1} ) \\ \\ = \frac{ \sqrt{3} }{2} + 3( \frac{4 \sqrt{5} + 4 }{4 \sqrt{5} + 4} ) \\ = \frac{3 \sqrt{3} }{2}$

hope this helps you please mark it as brainliest

Answer 2

Answer:

Sharon pls vaapis aa jao