Question

$Solve:{\text{ secx - 1 = }}\left( {\sqrt 2 - 1} \right)\tan x,{\text{ x}} \ne \left( {2n - 1} \right)\dfrac{\pi }{2},{\text{ }}n \in Z$
A. $2n\pi ,{\text{ }}\left( {4n + 1} \right)\dfrac{\pi }{2},{\text{ n}} \in {\text{Z}}$
B. $n\pi ,{\text{ }}\left( {4n + 1} \right)\dfrac{\pi }{2},{\text{ n}} \in {\text{Z}}$
C. $2n\pi ,{\text{ 2n}}\pi {\text{ + }}\dfrac{\pi }{4},{\text{ n}} \in {\text{Z}}$
D. $2n\pi ,{\text{ }}\left( {2n + 1} \right)\dfrac{\pi }{2},{\text{ n}} \in {\text{Z}}$

Answer 1

This is a trigonometry question. So one must know the identities related to it. The key observation to this question is the value $\left( {\sqrt 2 - 1} \right)$ can be substituted by $\tan \dfrac{\pi }{8}$. The other identities will also be used in this question such as $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$. With this we will be using some other trigonometric formulas. After simplifying this we will get the final answer.

Complete step-by-step answer:
The given equation is,
${\text{secx - 1 = }}\left( {\sqrt 2 - 1} \right)\tan x$
As it is known that $\tan \dfrac{\pi }{8} = \left( {\sqrt 2 - 1} \right)$
$\therefore$ The equation can be written as,
${\text{secx - 1 = tan}}\dfrac{\pi }{8}\tan x$
$\because$ $\tan x = \dfrac{{\sin x}}{{\cos x}}$
And $\sec x = \dfrac{1}{{\cos x}}$
$\therefore$ The equation becomes,
$\dfrac{1}{{\cos x}} - 1 = \dfrac{{\sin \dfrac{\pi }{8}\sin x}}{{\cos \dfrac{\pi }{8}\cos x}}$
On taking LCM in the LHS,
$\Rightarrow \dfrac{{1 - \cos x}}{{\cos x}} = \dfrac{{\sin \dfrac{\pi }{8}\sin x}}{{\cos \dfrac{\pi }{8}\cos x}}$
Multiplying both sides by $\cos x$,
$\Rightarrow 1 - \cos x = \dfrac{{\sin \dfrac{\pi }{8}\sin x}}{{\cos \dfrac{\pi }{8}}}$
Multiplying both sides by $\cos \dfrac{\pi }{8}$,
$\Rightarrow \cos \dfrac{\pi }{8} - \cos x\cos \dfrac{\pi }{8} = \sin \dfrac{\pi }{8}\sin x$
Adding both sides by $\cos x\cos \dfrac{\pi }{8}$
$\Rightarrow \cos \dfrac{\pi }{8} = \sin \dfrac{\pi }{8}\sin x + \cos x\cos \dfrac{\pi }{8}$
Or,
$\sin \dfrac{\pi }{8}\sin x + \cos x\cos \dfrac{\pi }{8} = \cos \dfrac{\pi }{8}$
$\because {\text{ }}\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\therefore$ Equation can written as,
$\cos \left( {x - \dfrac{\pi }{8}} \right) = \cos \dfrac{\pi }{8}$
$\Rightarrow \left( {x - \dfrac{\pi }{8}} \right) = 2n\pi \pm \dfrac{\pi }{8},{\text{ }}\forall {\text{ n}} \in {\text{Z}}$
When $\left( {x - \dfrac{\pi }{8}} \right) = 2n\pi + \dfrac{\pi }{8}$
$\Rightarrow x = 2n\pi + \dfrac{\pi }{4}$
And when $\left( {x - \dfrac{\pi }{8}} \right) = 2n\pi - \dfrac{\pi }{8}$
$\Rightarrow x = 2n\pi$
$\therefore$ $x = 2n\pi + \dfrac{\pi }{4}{\text{ , n}} \in {\text{Z}}$
And $x = 2n\pi ,{\text{ n}} \in Z$
Are the solutions of the given equation.

So, the correct answer is “Option C”.

Note: This is trigonometry-based question and in order to solve this the identities related to it must be known. The equation should be solved carefully. All the multiple and compound angle formulas should be known. After the final answer is found out it can be checked that whether it satisfies the original equation given in the question by simply substituting its value in the equation and if it does not satisfy the equation then the solution must be rechecked. The equation should be solved in accordance with the identities which would result in the correct solution. The value of n in this question can be any integer and for all such integers it satisfies the given equation. Calculations should be done carefully to avoid any mistake. Always try to solve the question step by step so that the wrong step can be determined and changed.