Question

If ${\tan }^2\alpha$ = $1-p^2$, then $\text{s}ec\alpha$ + ${\tan }^3\alpha$ $\text{c}osec\alpha$ =

Answer 1

${\tan }^2\alpha =1-p^2$

${\sec }^2\theta -{\tan }^2\theta =1$ (formula)

${\sec }^2\alpha =1+{\tan }^2\alpha \Rightarrow {\sec }^2\alpha =1+1-p^2=2-p^2$

$\sec \alpha +{\tan }^3\alpha co\sec \alpha =\frac{1}{\cos \alpha }+\frac{{\sin }^3\alpha }{{\cos }^3\alpha }=\frac{1}{\cos \alpha }+\frac{1-{\cos }^2\alpha }{{\cos }^3\alpha }[{\sin }^2\alpha +{\cos }^2\alpha =1]$

$=\frac{1}{\cos \alpha }+\frac{1}{{\cos }^3\alpha }-\frac{{\cos }^2\alpha }{{\cos }^3\alpha }=\frac{1}{\cos \alpha }+\frac{1}{{\cos }^3\alpha }-\frac{1}{\cos \alpha }={\sec }^3\alpha [\because {\sec }^2\alpha =2-p^2]$

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