Pleass do the sum quckly plz
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We have the values...
tan(π/3)=√3
tan(π/4)=1
tan(π/6)=1/√3
we can see that..
√3 , 1 , 1/√3
form a gp where common ratio r=1/√3
because.. if a is first term
a=√3
ar=(√3)(1/√3)
=√3/√3
=1
ar^2=ar.r
=(1)(1/√3)
=1/√3
tan(π/3)=√3
tan(π/4)=1
tan(π/6)=1/√3
we can see that..
√3 , 1 , 1/√3
form a gp where common ratio r=1/√3
because.. if a is first term
a=√3
ar=(√3)(1/√3)
=√3/√3
=1
ar^2=ar.r
=(1)(1/√3)
=1/√3
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