3
$\begingroup$

Solve $ \sec x- 1 = (\sqrt 2 - 1) \tan x $

Case 1)

Square both the sides and using $ \sec ^2 x = 1+ \tan^2 x. $ And solving the quadratic we get answer $\tan x = 1$ or $\tan x = 0$. Putting them back in also solves the equation.

Thus $x$ is either $n \pi $ or $x = n \pi + \frac{\pi}{4}, n \in \mathbb{Z}$

Case 2)

$ \sec x- 1 = (\sqrt 2 - 1) \tan x $

$ \frac{1-\cos x}{\cos x} = (\sqrt 2 - 1) \frac{\sin x}{\cos x} $

$ 2\sin^2 \frac{x}{2}= (\sqrt 2 - 1) 2 \sin \frac{x}{2} \cos \frac{x}{2}$

Thus solution is $\sin \frac{x}{2} = 0$ and $\tan \frac{x}{2} = \frac{\pi}{8}$

Thus $x$ is either $x = 2n \pi$ Or $x = 2n\pi + \frac{π}{4}$

These two answer are not same. So something is wrong.

Even though below question look similar to this one. None of the concepts in its answers really help this question. Different ways gives different results - solving $\tan 2a = \sqrt 3 $

$\endgroup$
2
  • 1
    $\begingroup$ Case 1): $n\pi$ is not a solution for $n$ odd. $\endgroup$ – Kavi Rama Murthy May 8 '19 at 8:41
  • 1
    $\begingroup$ $$\sqrt2-1=\dfrac{\sec x-1}{\tan x}=\csc x-\cot x$$ $$\iff\sqrt2+1=\csc x+\cot x$$ as $\csc^2x-\cot^2x=1$ $\implies\csc x=\sqrt2>0,\cot x=1>0$ So, $x$ in the first quadrant $$\implies x=2n\pi+\dfrac\pi4$$ $\endgroup$ – lab bhattacharjee May 8 '19 at 8:59
2
$\begingroup$

Squaring both sides of an equation can introduce extraneous solutions. Substitution demonstrates that if $n$ is odd, then $x = n\pi$ is not a solution as $-2 \neq 0$ and that $x = n\pi + \frac{\pi}{4}$ is not a solution as the LHS is negative while the RHS is positive. On the other hand, direct substitution shows that your second method produces correct solutions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.