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Question

If A+B+C=π/2, then show that
i. cotA+cotB+cotC=cotAcotBcotC
ii. tanAtanB+tanBtanC+tanCtanA=1.

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Solution

A+B+C=π2

cotA+cotB+cotC

simplifying in terms of sin and cos

=cosAsinA+cosBsinB+cosCsinC

=(cosAsinBsinc+cosBsinAsinc+cosCsinAsinB)sinAsinBsinC

=cos(A+B+C)+cosAcosBcosCsinAsinBsinC=cosAcosBcosCsinasinBsinC=cotAcotBcotC
tanAtanB+tanBtanC+tanCtanA

simplifying in terms of sin and cos

=sinAcosA.sinBcosB.sinCcosC+sinCcosC.sinAcosA

=sinAsinBcosC+sinBsinCcosA+sinCsinAcosBcosAcosBcosC

=cosAcosBcosCcos(A+B+C)cosAcosBcosC=1

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