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Help me with it please with steps

class 9 trigonometry ​

Help me with it please with steps class 9 trigonometry ​-example-1
User Kruschid
by
6.4k points

2 Answers

4 votes

Answer:


(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)

User BarretV
by
6.8k points
3 votes

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

User Rich MacDonald
by
7.2k points