Question

Modulus function : Let $f:R\to R$ be given by $f\left(x\right)=|x|$ for each $x\in R$. then $f\left(x\right)=|x|=x$, when $x\ge 0$
$=-x$, when $x<0$
The number of solution of the equation $|\cot x|=\cot x+\csc x;0\le x\le 2\pi$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$\cot x+\csc x=|\cot x|$ ... (1)

$⟹\cot x+\csc x=\pm \cot x$

$⟹\cot x+\csc x=\cot x\text{and/or}\cot x+\csc x=-\cot x$

$⟹\csc x=0\text{and/or}\csc x=-2\cot x$

It is impossible that $\csc x=0$ as $\csc x\in (-\infty ,-1]\cup [1,\infty )$

$⟹\csc x=-2\cot x$

$⟹\cos x=-\frac{1}{2}$

$⟹x=\frac{2\pi }{3}\text{or}\frac{4\pi }{3}$

For $x=\frac{2\pi }{3}$, $\cot x+\csc x=-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}=|\cot x|$ which satisfies equation (1)

For $x=\frac{4\pi }{3}$, $\cot x+\csc x=\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$ which does not satisfy equation (1)

Hence, only 1 solution exists.