If $tan^2θ = 1-a^2$, then the value of $secθ+ tan^3θ cosecθ$ is:
Answer & explanation
Correct answer: option 3
$secθ+ tan^3θ cosecθ$
= \(\frac{1}{cosθ}\) + \(\frac{sin3θ}{cos3θ}\) × \(\frac{1}{sinθ}\)
= \(\frac{1}{cosθ}\) [ 1 + \(\frac{sin2θ}{cos2θ}\) ]
= \(\frac{1}{cosθ}\) [ \(\frac{ cos2θ + sin2θ }{cos2θ}\) ]
= \(\frac{1}{cosθ}\) × \(\frac{ 1 }{cos2θ}\)
= \(\frac{ 1 }{cos3θ}\)
= sec3θ
As ,tan2θ = 1 - a2
sec2θ - 1 = 1 - a2
( tan2θ = sec2θ - 1 )
sec2θ = 2 - a2
sec3θ = $(2-a^2)^{\frac{3}{2}}$