# How do you verify: $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$?

Answer

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**Hint:**We have sum of two terms in the left-hand side of $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$. 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 solution:**

We have the sum of two terms in $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}$. We take the LCM of the denominators as the multiplications of those terms.

The equation becomes $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{{{\cos }^{2}}x+{{\left( 1+\sin x \right)}^{2}}}{\cos x\left( 1+\sin x \right)}$.

We break the square as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, ${{\left( 1+\sin x \right)}^{2}}=1+{{\sin }^{2}}x+2\sin x$.

The equation becomes $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{{{\cos }^{2}}x+{{\left( 1+\sin x \right)}^{2}}}{\cos x\left( 1+\sin x \right)}=\dfrac{{{\cos }^{2}}x+1+{{\sin }^{2}}x+2\sin x}{\cos x\left( 1+\sin x \right)}$.

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We get $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2+2\sin x}{\cos x\left( 1+\sin x \right)}$.

We take 2 common from the numerator to get $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}$.

We omit the common from both numerator and denominator and get

$\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}=\dfrac{2}{\cos x}$.

We use the inverse formula to get $\dfrac{1}{\cos x}=\sec x$. So, $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$.

Thus verified $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$.

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

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