if tan3Q. - 1 over tan Q -1. = A sec square theta + B tan square theta then a + b is equal to
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secθ+tan3θcosecθ
=secθ[secθsecθ+tan3θcosecθ]
=secθ[1+tan3θ⋅sinθcosθ]
=secθ[1+tan3θ×cotθ]
=1+tan2θ[1+tan2θ]
=[1+tan2θ]3/2
=[1−(1−a2)]3/2=(2−a2)3/2
[∵tan2θ=1−a2]
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