Question

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

Answer 1

Answer:

I think that the answer is 189/12.

Answer 2

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

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