## Different methods give different answers Solve $\sec x- 1 = (\sqrt 2 – 1) \tan x$

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

# Case 1)

Square both the sides and using $$\sec ^2 x = 1+ \tan^2 x.$$We get answer tan x = 1 or tan x = 0.

Thus X is either $$n \pi$$ Or $$X is n \pi + \frac{π}{4}$$

# case 2.

$$\sec x- 1 = (\sqrt 2 – 1) \tan x$$

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

$$2\sin^2 X/2= (\sqrt 2 – 1) 2 \sin X/2 \cos X/2$$

Thus solution is sin X/2 = 0 and tan x/2 = π/8

Thus X is either $$X = 2n \pi$$ Or $$X = 2n\pi + π/4$$

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 $\tan 2a = \sqrt 3$