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Question

Evaluate each of the following:

(i) tan-1tanπ3
(ii) tan-1tan6π7
(iii) tan-1tan7π6
(iv) tan-1tan9π4
(v) tan-1tan1
(v) tan-1tan2
(v) tan-1tan4
(v) tan-1tan12

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Solution

We know that

tan1tanθ=θ, -π2<θ<π2

(i) We have
tan-1tanπ3=π3
(ii) We have
tan-1tan6π7=tan-1tanπ-π7=tan-1tan-π7=-π7
(iii) We have

tan-1tan7π6=tan-1tanπ+π6=tan-1tanπ6=π6
(iv) We have

tan-1tan9π4=tan-1tan2π+π4=tan-1tanπ4=π4
(v) We have
tan-1tan1=1
(vi) We have

tan-1tan2=tan-1tan-π+2=2-π

(vii) We have

tan-1tan4=tan-1tan-π+4=4-π

(viii) We have

tan-1tan12=tan-1tan-4π+12=12-4π

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