Question

How do you simplify $\dfrac{1}{1+\sin x}+\dfrac{1}{1-\sin x}$?

Answer 1

Hint : For the given question we are given to simplify the equation $\dfrac{1}{1+\sin x}+\dfrac{1}{1-\sin x}$. For that we have to observe the equation that we could solve the equation by examining the equation. To solve the given question, we have moved the all the equation to one side and complete the summation with the help of LCM. Then we use the identity formula of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Complete step-by-step answer :
The given expression is $\dfrac{1}{1+\sin x}+\dfrac{1}{1-\sin x}$.
We simplify the expression taking the LCM of the denominators $1+\sin x$ and $1-\sin x$.
The LCM becomes $\left( 1+\sin x \right)\left( 1-\sin x \right)$

& \dfrac{1}{1+\sin x}+\dfrac{1}{1-\sin x} \\\ & =\dfrac{\left( 1-\sin x \right)+\left( 1+\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)} \\\ \end{aligned}$$ We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Therefore, $\left( 1+\sin x \right)\left( 1-\sin x \right)=1-{{\sin }^{2}}x$. The simplification of the numerator gives $$\left( 1-\sin x \right)+\left( 1+\sin x \right)=2$$ . The simplified form is $$\dfrac{\left( 1-\sin x \right)+\left( 1+\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}=\dfrac{2}{1-{{\sin }^{2}}x}$$ We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ which gives $1-{{\sin }^{2}}x={{\cos }^{2}}x$. Therefore, $$\dfrac{2}{1-{{\sin }^{2}}x}=\dfrac{2}{{{\cos }^{2}}x}=2{{\sec }^{2}}x$$ as we know $$\sec x=\dfrac{1}{\cos x}$$ . Therefore, the simplified form of $\dfrac{1}{1+\sin x}+\dfrac{1}{1-\sin x}$ is $$2{{\sec }^{2}}x$$ . **So, the correct answer is “ $$2{{\sec }^{2}}x$$ .”.** **Note** : We need to remember that the final terms are square terms. The identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\sin x\times \csc x=1$ are valid for any value of $x$. We also can change the form $$2{{\sec }^{2}}x=2\left( 1+{{\tan }^{2}}x \right)$$ as $${{\sec }^{2}}x=1+{{\tan }^{2}}x$$ .