What is the area of this figure?
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Ответ:

mandylo5418

29.03.2023

Mathematics

$\frac{13}{7}$

Step-by-step explanation:

Using the identities

cos x = $\frac{1}{secx}$ , sin²x = 1 - cos²x

tan²x = sec²x - 1

Given

$\frac{3cos^2A+5tan^2A}{4tan^2A-sin^2A}$

= $\frac{3(\frac{1}{sec^2A}+5(sec^2A-1) }{4(sec^2A -(1-cos^2A)}$

= $\frac{3.(\frac{1}{\sqrt{2})^2 } +5((\sqrt{2})^2-1) }{4((\sqrt{2})^2-1)-(1-(\frac{1}{\sqrt{2} })^2 }$

= $\frac{\frac{3}{2}+5(2-1) }{4(2-1)-(1-\frac{1}{2} )}$

= $\frac{\frac{3}{2} +5}{4-\frac{1}{2} }$

= $\frac{\frac{13}{2} }{\frac{7}{2} }$

= $\frac{13}{2}$ × $\frac{2}{7}$

= $\frac{13}{7}$

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