Question

Solve the following equations.
$|cot(2x-\frac{\pi }{2})|=\frac{1}{co{s}^{2}2x}-1.$

Answer 1

$|cot(2x-\frac{\pi }{2})|=\frac{1}{co{s}^{2}2x}-1...\left(1\right)$

$\because cot(2x-\frac{\pi }{2})=tan2x,\frac{1}{co{s}^{2}2x}=se{c}^2\left(2x\right)$

i-e put in eq(1)

$\left|tan2x\right|=se{c}^2\left(2x\right)-1$

$\left|tan2x\right|=ta{n}^2\left(2x\right)$ $[\because ta{n}^2\theta =se{c}^2\theta -1]$

We know square quantity only have

positive value

i.e.

$tan2x=ta{n}^2\left(2x\right)$

$tan2x-ta{n}^2\left(2x\right)=0$

$tan2x(1-tan(2x\left)\right)=\mathrm{0...}\left(2\right)$

Now solving the eq (2)

$\Rightarrow tan2x=0$ or $1-tan2x=0$

Here $x\to \infty$ Or $1=tan2x$

which is not possible Or $2x=ta{n}^{-1}\left(1\right)$

i.e $2x=\pi /4$

$x=\pi /8$