Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 13th Jul 2024
Total views: 346.2k
Views today: 6.46k
Verified
346.2k+ views
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 $x\ne \dfrac{\left( 4n-1 \right)\pi }{2},n\in \mathbb{Z}$.