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help me with it please with steps

class 9 trigonometry ​

Help Me With It Please With Stepsclass 9 Trigonometry class=

Sagot :

kristinaneupane21

secA=√2

secA=sec45

A=45°

3cos²45+5tan²45/ 4tan²A-sin²45

= 3(1/√3)²+5(1)² / 4(1)²-(1/√2)²

=3/2+5/4-1/2

=13/7

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]

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