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Question

tanA1cotA+cotA1tanA=(secAcosecA+1)


Solution

LHS=tanA1cotA+cotA1tanA=(sinA/cosA)(1cosAsinA)+(cosA/sinA)1sinAcosA

sinAcosA(sinAcosA)sinA+cosAsinA(cosAsinA)cosA

=sin2AcosA(sinAcosA)+cos2AsinA(cosAsinA)=sin3Acos3AcosAsinA(sinAcosA)

=(sinAcosA)(sin2A+cos2A+sinAcosA)cosAsinA(sinAcosA) [Using a3b3=(ab)(a2+b2+ab)]

=1+sinAcosAsinAcosA[sin2A+cos2A=1]

=1sinAcosA=sinAcosAsinacosA

=sec A cosec A+1

[1cosA=secA,1sinA=cosecA]

=RHS

Hence proved.


Mathematics
RD Sharma
Standard XI

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