Question

If ${\tan }^2\theta =1-a^2$, prove that $\sec \theta +{\tan }^3\theta cosec\theta =(2-a^2{)}^{3/2}$

Answer 1

$sec\theta +ta{n}^3\theta cosec\theta$

$=sec\theta \left[\frac{sec\theta +ta{n}^3\theta cosec\theta }{sec\theta }\right]$

$=sec\theta [1+ta{n}^3\theta .\frac{cos\theta }{sin\theta }]$

$=\sqrt{1+ta{n}^2\theta }[1+ta{n}^2\theta ]$

$=(1+ta{n}^2\theta {)}^{3/2}$

$=[1+(1-a^2){]}^{3/2}$

$=(2-a^2{)}^{3/2}$