Question

The ratio between $\frac{1+\cos \alpha }{{\sin }^2\alpha }$ and ${\left(\frac{1+\cos \alpha }{{\sin }^2\alpha }\right)}^2$ in that order is

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $3:1$

Answer 1

The correct option is A $1:1$

$\frac{1+\cos \alpha }{{\sin }^2\alpha }:{\left(\frac{1+\cos \alpha }{{\sin }^2\alpha }\right)}^2$

$\Rightarrow 1:\frac{1+\cos \alpha }{{\sin }^2\alpha }$

$\Rightarrow 1:\frac{1+\cos \alpha }{(1-{\cos }^2\alpha )}$

$\Rightarrow 1:\frac{1+\cos \alpha }{(1-\cos \alpha )(1+\cos \alpha )}$

$\Rightarrow 1:\frac{1}{(1-\cos \alpha )}$

$\Rightarrow 1-\cos \alpha :1$

$\Rightarrow \frac{1-\cos \alpha }{1}=1$

$\Rightarrow 1-\cos \alpha =1$

$\Rightarrow \cos \alpha =0$

$\Rightarrow \alpha =\frac{\pi }{2}$

$\therefore$ the ratio at $\alpha =\frac{\pi }{2}$ is $\frac{1+\cos \frac{\pi }{2}}{{\sin }^2\frac{\pi }{2}}:{\left(\frac{1+\cos \frac{\pi }{2}}{{\sin }^2\frac{\pi }{2}}\right)}^2$

$=\frac{1}{1}:\frac{1}{1}=1:1$