Question

The value of $\frac{\cos ({90}^o-A)}{1+\sin ({90}^o-A)}+\frac{1+\sin ({90}^o-A)}{\cos ({90}^o-A)}$ is equal to :

• A. $\frac{2}{\cos A}$
• B. $\frac{2}{\sin A}$
• C. $\frac{2}{\sec A}$
• D. $\frac{2}{\text{cosec}A}$

Answer 1

Answer: (B)

$\frac{2}{\sin A}$

Given that:

$\frac{\cos ({90}^o-A)}{1+\sin ({90}^o-A)}+\frac{1+\sin ({90}^o-A)}{\cos ({90}^o-A)}$

$=\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=\frac{{\sin }^{2}A+(1+\cos A{)}^2}{\sin A(1+\cos A)}$

$=\frac{{\sin }^{2}A+{\cos }^{2}A+2\cos A+1}{\sin A(1+\cos A)}=\frac{2+2\cos A}{\sin A(1+\cos A)}=\frac{2}{\sin A}$