Question

Prove that $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$?

Answer 1

Hint: We have sum of two terms in the left-hand side of $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$. We take the LCM of the denominators. Then we add them in the denominators. Then we use the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to simplify the numerator and also use ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Complete step-by-step answer:
We have the sum of two terms in $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$. We take the LCM of the denominators as the multiplications of those terms.
The equation becomes
$\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\left( 1+\sin A \right)}^{2}}}{\cos A\left( 1+\sin A \right)}$.
We break the square as
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, ${{\left( 1+\sin A \right)}^{2}}=1+{{\sin }^{2}}A+2\sin A$.
The equation becomes
$\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\left( 1+\sin A \right)}^{2}}}{\cos A\left( 1+\sin A \right)}=\dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{\cos A\left( 1+\sin A \right)}$.
We know that
${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We get $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\cos A\left( 1+\sin A \right)}$.
We take 2 common from the numerator to get
$\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\cos A\left( 1+\sin A \right)}$.
We omit the common from both numerator and denominator and get
$\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\cos A\left( 1+\sin A \right)}=\dfrac{2}{\cos A}$.
We use the inverse formula to get
$\dfrac{1}{\cos x}=\sec x$. So, $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2}{\cos A}=2\sec A$.
Thus verified $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.

Note: It is important to remember that the condition to eliminate the $\left( 1+\sin A \right)$ from both denominator and numerator is $\left( 1+\sin A \right)\ne 0$. No domain is given for the variable $A$. The value of $\sin A\ne -1$ is essential. The simplified condition will be $A\ne \dfrac{\left( 4n-1 \right)\pi }{2},n\in \mathbb{Z}$.