Math, asked by debarshishivam123, 11 hours ago

y= 1+cos2x+8sin²x/ sin2x
find y min when x belongs to (0, pi/2)

Answers

Answered by avabooleav

0

Answer:

Step-by-step explanation:

f(x)=

sin2x

1+cos2x+8sin

2

x

=

sinxcosx

cos

2

x+4sin

2

x

=cotx+4tanx

2

cotx+4tanx

≥

4

⟹f(x)≥4

Minimum value of f(x) is 4

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