If Cos A =2/5,find the value of 4+tan²A
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Answered by
1
cosA=2/5
base/hypotenuse=2/5
let base=2k and hypotenuse=5k
by Pythagoras theorem
perpendicular=√21k
now tanA=√21/2
now 4+ tanA.tanA
= 4+(√21/2)(√21/2)
=4+21/4
=(16+21)/4
=37/4
base/hypotenuse=2/5
let base=2k and hypotenuse=5k
by Pythagoras theorem
perpendicular=√21k
now tanA=√21/2
now 4+ tanA.tanA
= 4+(√21/2)(√21/2)
=4+21/4
=(16+21)/4
=37/4
Answered by
0
cos A=2/5=b/h
p^2=h^2-b^2
=(5)^2-(2)^2
=25-4
p=√21
4+tan^2 A
=4+(p/b)^2
=4+(√21/2)^2
=4+21/4
=16+21/4 (here 4 is L.C.M)
=37/4
p^2=h^2-b^2
=(5)^2-(2)^2
=25-4
p=√21
4+tan^2 A
=4+(p/b)^2
=4+(√21/2)^2
=4+21/4
=16+21/4 (here 4 is L.C.M)
=37/4
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