Question

$\frac{5 { \cos }^{2} 60 + 4 { \sec }^{2} 30 - { \tan }^{2}45 } { {sin}^{2} 30 + {sin}^{2}60 }$
​

Answer 1

Step-by-step explanation:

$\frac{5 { \cos }^{2} 60 + 4 { \sec }^{2} 30 - { \tan }^{2}45 } { {sin}^{2} 30 + {sin}^{2}60 } \\ = \frac{ 5{( \cos(60) )}^{2} + 4 {( \sec(30) )}^{2} - {( \tan(45)) }^{2} }{ { (\sin(30) )}^{2} + { (\sin(60) )}^{2} } \\ = \frac{5 {( \frac{1}{2} )}^{2} + 4 {( \frac{2}{ \sqrt{3} } )}^{2} - {(1)}^{2} }{ ({ \frac{1}{2}) }^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} } \\ = \frac{5 ( \frac{1}{4}) + 4( \frac{4}{3} ) - 1}{ \frac{1}{4} + \frac{3}{4} } \\ = \frac{ \frac{5}{4} + \frac{16}{3} - 1}{ \frac{1 + 3}{4} } \\ = \frac{ \frac{15 + 64}{12} - 1 }{ \frac{4}{4} } \\ = \frac{ \frac{79 - 12}{12} }{1} \\ = \frac{67}{12}$

Answer 2

Answer:

Hope this will help you. Plz mark me as brainliest.