Question

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

Answer 1

Hint: Here we have to simplify the value given. Firstly we will take the LCM of the two values given. Then we will simplify the values obtained by using the Pythagorean identity given as ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ . Finally by using the relation between the tangent, sin and cosine we will get our desired answer.

Complete Solution
We have to simplify the value given below:
$\dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}$….$\left( 1 \right)$
Now as we have two fraction values we will take the LCM of the above value as follows:
$\Rightarrow \dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}=\dfrac{\sin x\left( 1-\sin x \right)-\sin x\left( 1+\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ where \[a=1\] and $b=\sin x$ above we get,
$\Rightarrow \dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}=\dfrac{\sin x-{{\sin }^{2}}x-\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x}$
Canceling out the common terms in the numerator we get,
$\Rightarrow \dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}=\dfrac{-2{{\sin }^{2}}x}{1-{{\sin }^{2}}x}$….$\left( 2 \right)$
Now as we know Pythagorean identity is given as follows:
${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Take the cosine term on the right side so,
$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$
Use above value in equation (2) as follows:
$\Rightarrow \dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}=\dfrac{-2{{\sin }^{2}}x}{{{\cos }^{2}}x}$
Next as $\tan A=\dfrac{\sin A}{\cos A}$ using it above,
$\Rightarrow \dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}=-2{{\tan }^{2}}x$
So we got the answer as $-2{{\tan }^{2}}x$ .

Hence on simplifying $\dfrac{\sin x}{1+\sin x}-\dfrac{\sin x}{1-\sin x}$ we got the answer as $-2{{\tan }^{2}}x$ .

Note:
While solving such types of problems, always simplify the value by taking the LCM (Least Common Multiple) which in this case is taken as the product of the two denominator values. These kinds of questions can be asked as a proving problem. Then it is easy to approach the solution.