Math, asked by ruchi12345, 9 months ago

Find the value of :- cot^2 30°+8sin^2 45°+3/2sec^2 30°+2cos^2 90° upon 2sec60°+3cosec30°-7/3tan^2 60°​

Answers

Answered by prathamsaxena21
4

Answer:

I think that the answer is 189/12.

Answered by isafsafiya
8

Answer:

 9\\  \\

Given:-

 \frac{ { \cot}^{2} 30 + 8 {sin}^{2} 45 +  \frac{3}{2}  {sec}^{2} 30 + 2 {cos}^{2} 90}{2 \: sec \: 60 + 3cosec \: 30 -  \frac{7}{3} {tan}^{2}60  }  \\  \\

To find:-

  • value of this quetion

Solution:-

as we all know

sin \: 30 =  \frac{1}{2}  \\ sin \: 45 =  \frac{1}{ \sqrt{2} }  \\ sec \: 30 =   \frac{2}{ \sqrt{3} }  \\ cot 30 =  \sqrt{3}  \\ cos \: 90 = 0 \\ sec \: 60 = 2 \\ cosec \: 30 = 2 \\ tan \: 60 =  \sqrt{3}

now put all the value in quetion

we get

 \frac{ { \sin }^{2} 30 + 8 {sin}^{2} 45 +  \frac{3}{2}  {sec}^{2} 30 + 2 {cos}^{2} 90}{2 \: sec \: 60 + 3cosec \: 30 -  \frac{7}{3} {tan}^{2}60  }  \\  \\  \frac{ { \frac{1}{2} }^{2}  + 8 \times ( { \sqrt{ </p><p>√3}^{2}  +  \frac{3}{2} \times  ({ \frac{2}{ \sqrt{3} }) }^{2}  + 2 \times 0 }{2 + 3 \times 2 -  \frac{7}{3}  \times  {  \sqrt{3}  }^{2} }  \\  \\  \frac{ 3} + 8 \times  \frac{1}{2}  +  \frac{3}{2}   \times  \frac{4}{3}  + 0}{2 + 6 -  \frac{7}{3}  \times 3}  \\  \\   \frac{ 3+ 4 + 2 }{8 - 7}  \\  \\  \frac{ 3 + 6}{1}  \\  \\  3+ 6 \\  \\ 9

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