Math, asked by sharon9876, 9 months ago

Help me guys...Pls....​

Attachments:

Answers

Answered by ITzBrainlyGuy
1

\huge{\mathcal{\underline\green{Question}}}

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

\huge{\mathrm{\underline\pink{answer}}}

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

Answered by vanshg28
0

Answer:

Sharon pls vaapis aa jao

Similar questions