English, asked by durgaprasadmaths, 10 months ago

limit x tends to π/2 x/cosecx​

Answers

Answered by YashashwiRawat
1

Explanation:

y=x->π/2[(cosecx - 1)/(π/2 -x)²]

This is of the form 0/0.

y=x->π/2[(-cosecxcotx)/[2(π/2 - x)(-1)]

Cancelling -1

y=x->π/2[(cosecxcotx)/(π-2x)]

y=x->π/2[((1/sinx)(cosx/sinx))/(π-2x)]

y=x->π/2[(cosx)/(sin²x(π-2x))]

y=x->π/2[(sin(π/2 - x)/(sin²x.(π-2x))]

y=x->π/2[(sin(π/2 - x))/(2sin²x(π/2 - x))]

Ltx->π/2[(sin(π/2 - x))/(π/2 - x)]=1

y=1/(2sin²(π/2))

y=1/(2(1)²)

y=1/2

Hope this helps.........

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