Question

5sin²30 +cos²45 - 4tan²30÷ 2sìn30× cos30+ tan45⁰

Answer 1

Step-by-step explanation:

your answer

plzzz mark brainliest...

Answer 2

Given :

$\frac{5 {sin}^{2} 30 + {cos}^{2} 45 - 4 {tan}^{2} 30}{2sin30 \times cos30} + tan \: 45$

Values :

$sin \: \: 30 = \frac{1}{2} \\ \\ cos \: \: 45 = \frac{1}{ \sqrt{2} } \\ \\ tan \: \: 30 = \frac{1}{ \sqrt{3} } \\ \\ cos \: 30 = \frac{ \sqrt{3} }{2} \\ \\ tan \: 45 = 1$

Solution :

$= \frac{5 {sin}^{2}30 + {cos}^{2} 45 - 4 {tan}^{2}30 }{2sin \: 30 \times cos \: 30 + tan \: 45} \\ \\ = \frac{5 {( \frac{1}{2} )}^{2} + ( { \frac{1}{ \sqrt{2} } }) }^{2} - 4( { \frac{1}{ \sqrt{3} }) }^{2} }{2( \frac{1}{2} )( \frac{ \sqrt{3} }{2}) + 1 } \\ \\ = \frac{5 \times \frac{1}{4} + \frac{1}{2} - 4 \times \frac{1}{3} }{ \frac{ \sqrt{3} }{2} + 1}$

By finding the LCM of denominator i. e. 12 we get ,,,,

$= \frac{ \frac{15}{12} + \frac{6}{12} - \frac{16}{12} }{ \frac{ \sqrt{3} + 2 }{2} } \\ \\ = \frac{5}{12} \times \frac{2}{2 + \sqrt{3} } \\ \\ = \frac{5}{6(2 + \sqrt{3)} } \\ \\ = \frac{5}{12 + 6 \sqrt{3} }$

By rationalizing the denominator ,,,, we get ,,,,

$= \frac{5}{12 + 6 \sqrt{3} } \times \frac{12 - 6 \sqrt{3} }{12 - 6 \sqrt{3} } \\ \\ = \frac{60 - 30 \sqrt{3} }{ {(12)}^{2} - {(6 \sqrt{3)} }^{2} } \\ \\ = \frac{30(2 - \sqrt{3} )}{144 - 36 \times 3} \\ \\ = \frac{30(2 - \sqrt{3)} }{144 - 108} \\ \\ = \frac{30(2 - \sqrt{3)} }{36} \\ \\ = \frac{15(2 - \sqrt{3} )}{18}$

Hope it's helpful .....

Please mark it as Brainliest ..... ❤