Find the value of tan13π/12
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tan(A - B) = [tan(A) - tan(B)]/[1 + tan(A)tan(B)].
Thus:
tan(13π/12)
= tan(4π/3 - π/4), from the hint
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)], from the above formula
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)], by rationalizing
= -(1 - √3)^2/(1 - 3), via difference of squares
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
Thus:
tan(13π/12)
= tan(4π/3 - π/4), from the hint
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)], from the above formula
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)], by rationalizing
= -(1 - √3)^2/(1 - 3), via difference of squares
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
PalakGusain:
,sorry but your answer is wrong
how..??
answer.... is √3-1/√3+1
yaaa .. my fault .. because i am using the repeated to rationalizing . stop before the rationalize.
Answered by
1
tan13π/12=.2679 this is your answer I hope you will be satisfied
no am not
wrong method
you told me find the value so i find the value
Oho
in which classs???
of u
i am class 12 and u
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