Question

5cos^260+4sec^230-tan^245/sin^230+cos^230

Answer 1

Answer:

67/12

Step-by-step explanation:

((5*((1/2)^2))+(4(2/(3^(1/2))^2))-(1^2))/(((1/2)^2)+((3^(1/2))^2))

=((5/4)+((4*4)/3)-(1))/((1/4)+(3/4))

=(((15+64-12)/12)/(4/4))

=(67/12)/1

=67/12

If there is any doubt related to my solution , please ask it in the comment section .

Answer 2

$\huge{\star{\orange{\boxed{\pink{\boxed{\underline{\mathbb{\red{ANSWER:-}}}}}}}}}$

$\huge\tt\purple{\frac{5 { \cos }^{2} 60° + 4 { \sec }^{2} 30° - { \tan }^{2} 45°}{{ \sin }^{2}30° + { \cos }^{2} 30°} }$

$\longrightarrow$$\huge\tt\purple{\frac{5 { (\frac{1}{2}) }^{2} + 4 { (\frac{2}{ \sqrt{3} }) }^{2} - {(1)}^{2} }{ {( \frac{1}{2}) }^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} } }$

$\longrightarrow$$\huge\tt\purple{\frac{5( \frac{1}{4}) + ( \frac{16}{3}) - 1 }{ \frac{1}{4} + \frac{3}{4} } }$

$\longrightarrow$$\huge\tt\purple{\frac{ \frac{15 + 64 - 12}{12} }{ \frac{4}{4} } }$

$\longrightarrow$$\huge\tt\purple{\frac{67}{12}}$