We have,
cosB= BaseHypotenuse=13
So, we draw a triangle ABC, right angled at C such that
Base =BC=1 unit and, Hypotenuse =AB=3 units.
By pythagoras theorem, we have
AB2=BC2+AC2
⇒32=12+AC2
⇒AC2=9−1=8
⇒AC=√8=2√2
When we consider the t-ratios of ∠B, we have
Base =BC=1, Perpendicular =AC= 2√2, and Hypotenuse =AB=3
∴sinB=PerpendicularHypotenuse=2√23
tanB=PerpendicularBase=2√21=2√2
cosecB=HypotenusePerpendicular=32√2
secB=HypotenuseBase=31=3
cotB=BasePerpendicular=12√2
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