If tan2θ = 1-a2 find the value of secθ + tan3θ cosecθ.

We have to find secθ + tan³θ cosecθ

= secθ + tan²θ×\(\frac{sinθ}{cosθ}\)×cosecθ

= secθ + tan²θsecθ

= secθ (1+tan²θ)  (secθ=\(\sqrt {1+tan^2θ}\))

= \(\sqrt {1+tan^2θ}\)(1+tan²θ)

⇒ \( {(1+tan^2θ)}^{\frac{1}{2}+1} \)

⇒ \( {(1+tan^2θ)}^{\frac{3}{2}} \)

⇒ \( {(1+1-a^2)}^{\frac{3}{2}} \)

=\( {(2 - a^2)}^{\frac{3}{2}} \)