Question

How do you verify $\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\tan x\sec x$?

Answer 1

Hint: In this problem we need to verify the given equation. For this we will consider the left-hand side part and we will simplify that part. On the left-hand side, we can observe that the two fractions are in subtraction. We know that $\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$. So, we will compare the left-hand side part with $\dfrac{a}{b}-\dfrac{c}{d}$ and write the values of $a$, $b$, $c$, $d$. Now we will find the all the values which are necessary for the formula $\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$ which are $ad$, $bc$, $bd$. After calculating all the necessary values, we will substitute them in the formula and simplify them to get the required result.

Complete step by step answer:
Given equation $\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\tan x\sec x$.
In the above equation the left-hand side part is
$LHS=\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}$
Comparing the above equation with $\dfrac{a}{b}-\dfrac{c}{d}$, then we will get the values
$a=d=1+\sin x$, $b=c=1-\sin x$.
Now the value of $ad$ will be
$\begin{align}
  & ad=\left( 1+\sin x \right)\left( 1+\sin x \right) \\
 & \Rightarrow ad={{\left( 1+\sin x \right)}^{2}} \\
\end{align}$
Applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ in the above equation, then we will get
$\Rightarrow ad=1+{{\sin }^{2}}x+2\sin x$
Now the value of $bc$ will be
$\begin{align}
  & bc=\left( 1-\sin x \right)\left( 1-\sin x \right) \\
 & \Rightarrow bc={{\left( 1-\sin x \right)}^{2}} \\
\end{align}$
Applying the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ in the above equation, then we will get
$\Rightarrow bc=1+{{\sin }^{2}}x-2\sin x$
Now the value of $bd$ will be
$bd=\left( 1-\sin x \right)\left( 1+\sin x \right)$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, then we will have
$\Rightarrow bd=1-{{\sin }^{2}}x$
From the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, the value of $bd$ will be
$\Rightarrow bd={{\cos }^{2}}x$.
Now substituting all the values, we have in the formula $\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$, then we will get
$\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=\dfrac{\left( 1+{{\sin }^{2}}x+2\sin x \right)-\left( 1+{{\sin }^{2}}x-2\sin x \right)}{{{\cos }^{2}}x}$
Simplifying the above equation, then we will get
$\begin{align}
  & \dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=\dfrac{1+{{\sin }^{2}}x+2\sin x-1-{{\sin }^{2}}x+2\sin x}{{{\cos }^{2}}x} \\
 & \Rightarrow \dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=\dfrac{4\sin x}{{{\cos }^{2}}x} \\
\end{align}$
We can write ${{\cos }^{2}}x$ as $\cos x\times \cos x$, now the above equation is modified as
$\Rightarrow \dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\times \dfrac{\sin x}{\cos x}\times \dfrac{1}{\cos x}$
We have trigonometric formulas $\dfrac{\sin x}{\cos x}=\tan x$, $\dfrac{1}{\cos x}=\sec x$. Substituting these values in the above equation, then we will get
$\Rightarrow \dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\tan x\sec x$
Hence verified.

Note: We can also follow another method that is multiplying the term $\dfrac{1+\sin x}{1-\sin x}$with $\dfrac{1+\sin x}{1+\sin x}$ and the term $\dfrac{1-\sin x}{1+\sin x}$ with $\dfrac{1-\sin x}{1-\sin x}$ and apply all the trigonometric formula we have and simplify the equation to get the required result.