Given,tan θ = 1/√5In ∆ ABC,AC2 = AB2 + BC2AC2 = 12 + (√5)2AC2 = 1 + 5 = 6AC = √6We have,cosec θ = AC/AB = √6/1sec θ = AC/BC = √6/√5Now,

Q. 83.7( 6 Votes )

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Answer :

Given,

tan θ = 1/√5

In ∆ ABC,

AC² = AB² + BC²

AC² = 1² + (√5)²

AC² = 1 + 5 = 6

AC = √6

We have,

cosec θ = AC/AB = √6/1

sec θ = AC/BC = √6/√5

Now,

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