Question

${\tan }^2\alpha =1-p^2$ then, $\sec \alpha +{\tan }^3\alpha \text{cosec}\alpha =$

• A. $(2+p^2{)}^{\frac{3}{2}}$
• B. $(1+p^2{)}^{\frac{3}{2}}$
• C. $(2-p^2{)}^{\frac{3}{2}}$
• D. $(1-p^2{)}^{\frac{3}{2}}$

Answer 1

The correct option is C $(2-p^2{)}^{\frac{3}{2}}$

$\sec \alpha +{\tan }^3\alpha \text{cosec}\alpha =\frac{1}{\cos \alpha }+\frac{{\sin }^2\alpha }{{\cos }^3\alpha }=\frac{{\cos }^2\alpha +{\sin }^2\alpha }{{\cos }^3\alpha }=\frac{1}{{\cos }^3\alpha }={\sec }^3\alpha$

$=(1+{\tan }^2\alpha {)}^{\frac{3}{2}}=(2-p^2{)}^{\frac{3}{2}}$