Question

If \[\sqrt 2 \sec x + \tan x = 1\], then the value of \[x\] is
A.\[2n\pi + \dfrac{\pi }{3}\]
B.\[2n\pi - \dfrac{\pi }{4}\]
C.\[2n\pi + \dfrac{\pi }{6}\]
D.\[2n\pi + \dfrac{\pi }{{12}}\]

Answer 1

Hint: Here we will use the trigonometric properties. First, we will convert the given equation in terms of the sin and cos function and simplify it. Then we will put the value of the constants in the equation in terms of the trigonometric function. We will them simplify the equation to get the value of \[x\].

Complete step-by-step answer:
Given equation is \[\sqrt 2 \sec x + \tan x = 1\].
We will write the given equation in terms of the sine and cosine functions. We know that secant function is equal to the reciprocal of the cos function and tangent function is equal to the ratio of the sine to the cosine function. Therefore, we get
 \[ \Rightarrow \sqrt 2 \dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}} = 1\]
Now we will take the cos function common in the denominator and take it to the other side of the equation, we get
\[ \Rightarrow \dfrac{{\sqrt 2 + \sin x}}{{\cos x}} = 1\]
\[ \Rightarrow \sqrt 2 + \sin x = \cos x\]
Now we will simplify the above equation, we get
\[ \Rightarrow \cos x - \sin x = \sqrt 2 \]
Dividing both sides by \[\sqrt 2 \], we get
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }}\cos x - \dfrac{1}{{\sqrt 2 }}\sin x = 1\]
Now we will put the value of the constants in the equation in terms of the trigonometric functions.
We know that \[\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\] and \[\cos 2n\pi = 1\].
Now by putting these values in the above equation, we get
\[ \Rightarrow \cos \dfrac{\pi }{4}\cos x - \sin \dfrac{\pi }{4}\sin x = \cos 2n\pi \]
Now using the property of the trigonometry \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\], we get
\[ \Rightarrow \cos \left( {\dfrac{\pi }{4} + x} \right) = \cos 2n\pi \]
Now by cancelling the cos function from both sides of the equation, we get
\[ \Rightarrow \dfrac{\pi }{4} + x = 2n\pi \]
Now by solving this we will get the value of \[x\].
Subtracting \[\dfrac{\pi }{4}\] from both the sides, we get
\[ \Rightarrow x = 2n\pi - \dfrac{\pi }{4}\]
Hence the value of \[x\] is \[2n\pi - \dfrac{\pi }{4}\].
So, option B is the correct option.

Note: We know that there are six basic trigonometric functions and they are sine, cosine, tangent, cosecant, secant and cotangent. Also, cosecant, secant and cotangent are reciprocal functions of sine, cosine and tangent respectively. While solving trigonometric equations we should always write the trigonometric functions in terms of sine and cosine function. This makes it easier for us to solve the given equation. Every trigonometric function is a periodic function that means that they repeat their value after a certain interval. These intervals are the multiples of \[2\pi \].