We have,
cosecA= HypotenusePerpendicular=√101
So, we draw a triangle ABC, right angled at B such that
Perpendicular =BC=1 unit and, Hypotenuse =AC= √10 units.
By Pythagoras theorem, we have
AC2=AB2+BC2
⇒(√10)2=AB2+12
⇒AB2=10−1=9
⇒AB=√9=3
When we consider the trigonometric ratios of ∠ A , we have
Base =AB=3, Perpendicular =BC=1, and Hypotenuse =AC= √10
∴sinA=PerpendicularHypotenuse=1√10
cosA=BaseHypotenuse=3√10
tanA=PerpendicularBase=13
secA=HypotenuseBase=√103
and, cotA=BasePerpendicular=31=3
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