Question

Help me with it please with steps

class 9 trigonometry ​

Answer 1

`(13)/(7)`

Explanation:

Using the identities

cos x =
`(1)/(secx)` , sin²x = 1 - cos²x

tan²x = sec²x - 1

Given

`(3cos^2A+5tan^2A)/(4tan^2A-sin^2A)`

=
`(3((1)/(sec^2A)+5(sec^2A-1) )/(4(sec^2A -(1-cos^2A))`

=
`(3.((1)/(√(2))^2 ) +5((√(2))^2-1) )/(4((√(2))^2-1)-(1-((1)/(√(2) ))^2 )`

=
`((3)/(2)+5(2-1) )/(4(2-1)-(1-(1)/(2) ))`

=
`((3)/(2) +5)/(4-(1)/(2) )`

=
`((13)/(2) )/((7)/(2) )`

=
`(13)/(2)` ×
`(2)/(7)`

=
`(13)/(7)`

Answer 2

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