Answer:
[tex]\frac{13}{7}[/tex]
Step-by-step explanation:
Using the identities
cos x =[tex]\frac{1}{secx}[/tex] , sin²x = 1 - cos²x
tan²x = sec²x - 1
Given
[tex]\frac{3cos^2A+5tan^2A}{4tan^2A-sin^2A}[/tex]
= [tex]\frac{3(\frac{1}{sec^2A}+5(sec^2A-1) }{4(sec^2A -(1-cos^2A)}[/tex]
= [tex]\frac{3.(\frac{1}{\sqrt{2})^2 } +5((\sqrt{2})^2-1) }{4((\sqrt{2})^2-1)-(1-(\frac{1}{\sqrt{2} })^2 }[/tex]
= [tex]\frac{\frac{3}{2}+5(2-1) }{4(2-1)-(1-\frac{1}{2} )}[/tex]
= [tex]\frac{\frac{3}{2} +5}{4-\frac{1}{2} }[/tex]
= [tex]\frac{\frac{13}{2} }{\frac{7}{2} }[/tex]
= [tex]\frac{13}{2}[/tex] × [tex]\frac{2}{7}[/tex]
= [tex]\frac{13}{7}[/tex]
