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Question

If tan2θ=1a2, prove that secθ+tan3θcosecθ=(2a2)3/2

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Solution

We have

secθ+tan3θcosecθ

=secθ[secθ+tan3θcosecθsecθ]

=secθ[1+tan3θcosθsinθ]

=secθ[1+tan3θ×cotθ]

=1+tan2θ[1+tan2θ]

=[1+tan2θ]3/2

=[1(1a2)]3/2=(2a2)3/2

[tan2θ=1a2]

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