tan² 0 = 3 find all trigonometry rations
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tanA=1/
3
,
In right ΔABC
tanA=BC/AB=1/
3
,
So BC=1 and AB=
3
.
By Pythagoras theorem,
AC=
(AB
2
+BC
2
)
=
(
3
)
2
+(1)
2
=
(3+1)
=
4
=2.So,
sinA=BC/AC=1/2
cosA=AB/AC=
3
/2
cotA=1/tanA=
3
secA=1/cosA=2/
3
cosecA = 1/sinA=2/1=2.
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