Question

plz help me at trigonometry ​

Answer 1

Hey Pretty Stranger!

Given :

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

We know that :

$\bullet \sf \: \cos{60}^{ \circ} = \dfrac{1}{2}$

$\bullet \sf \: \tan{45}^{ \circ} = 1$

$\bullet \sf \: \sec{30}^{ \circ} = \dfrac{2}{ \sqrt{3} }$

So,

→ 5 × (½)² + 4 × (2/√3)² - (1)²/ 1

→ 5/4 + 4(4/3) - 1

→ 5/4 + 16/3 - 1

→ 15 + 64 - 12/12

→ 79 - 12/12

→ 67/12

Answer 2

$\frac{5 { \cos(60) }^{2} + 4 { \sec(30) }^{2} - { \tan(45) }^{2} }{ { \sin(30) }^{2} + { \cos(30) }^{2} } \\ = \frac{5 { (\frac{1}{ \sqrt{2} } )}^{2} + 4 { (\frac{2}{ \sqrt{3} }) }^{2} - {1}^{2} }{ { (\frac{1 }{ \sqrt{2}})}^{2} + { (\frac{ \sqrt{3} }{2} }^{2} ) } \\ = \frac{5 \times \frac{1}{2} + 4 \times \frac{4}{3} - 1 }{ \frac{1}{2} + \frac{3}{4} } \\ = \frac{ \frac{5}{2} + \frac{16}{3} - 1 }{ \frac{2 + 3}{4} } \\ = \frac{ \frac{15 + 32 - 6}{6} }{ \frac{5}{4} } \\ = \frac{ \frac{41}{6} }{ \frac{5}{4} } \\ = \frac{41}{6} \times \frac{4}{5} \\ = \frac{164}{30} \\ = \frac{82}{15}$