Question

$A=\frac{x-1}{x+1}$, then the value of $A-\frac{1}{A}$ is:

• A. $$\frac{-4(2x-1)}{x^2-1}$$
• B. $${x^2-1}{-4(2x-1)}$$
• C. $${x^2-1}{-4(2x+1)}$$
• D. $${-4x}{x^2-1}$$

Answer 1

Answer:

$${-4x}{x^2-1}$$

Given:

$A=\frac{x-1}{x+1}$

Formula used:

$(a+b)^2=a^2+2ab+b^2$

$(a^2-b^2)=(a-b)(a+b)$

Calculation:

$A-\frac{1}{A}$

Put the value of $A=\frac{x-1}{x+1}$ in the question

⇒ $\frac{(x-1)}{(x+1)}-\frac{(x+1)}{(x-1)}$

⇒$\frac{(x-1)\times{(x+1)}-(x+1)\times{(x+1)}}{x^2-1}$

⇒ $\frac{-4x}{x^2-1}$

∴ Correct answer is  $\frac{-4x}{x^2-1}$

Short trick:

Put the value of x = 2 

So,

⇒ $A = \frac{1}{3}$

According to the question,

$A-\frac{1}{A}$

⇒$\frac{1}{3}-3$

⇒ $\frac{-8}{3}$

Then check the option you get the answer 

Put the value in option (D)

⇒ $\frac{-4x}{x^2-1}$

⇒ $\frac{(-4\times{2})}{(4-1)}$

⇒$\frac{-8}{3}$

Correct answer is  $\frac{-4x}{x^2-1}$