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Question

(i) Evaluate (1+tanθ+secθ)(1+cotθcosecθ)
(ii) Prove that tanAsinAtanA+sinA=secA1secA+1

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Solution

(ii)

tanAsinAtanA+sinA=secA1secA+1

Taking LHS

tanAsinAtanA+sinA
=sinAcosAsinAsinAcosA+sinA
=sinAsecAsinAsinAsecA+sinA
=secA1secA+1


LHS=RHS

Hence proved.

(i)

(1+tanA+secA)(1+cotAcosecA)
(1+sinAcosA+1cosA)(1+cosAsinA1sinA)
=(cosA+sinA+1cosA)(cosA+sinA1sinA)
=(cosA+sinA)21sinAcosA
=sin2A+cos2A+2sinAcosA1sinAcosA
=1+2sinAcosA1sinAcosA

=2


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