Question

The most general solutions of the equation \[{\sec ^2}x = \sqrt 2 (1 - {\tan ^2}x)\] are given by
A. \[n\pi \pm \dfrac{\pi }{4}\]
B. \[2n\pi \pm \dfrac{\pi }{4}\]
C. \[n\pi \pm \dfrac{\pi }{8}\]
D. None of these

Answer 1

Hint: We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]
The general quadratic equation is of the form \[a{x^2} + bx + c = 0\] where a,b and c are constants and x is a variable.
Using the quadratic formula we can find the value of variable \[x\] as \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step answer:
When the arbitrary constant of the general solution takes some unique value, then the solution becomes the particular solution of the equation.
By using the boundary conditions (also known as the initial conditions) the particular solution of an equation is obtained.
So, to obtain a particular solution, first of all, a general solution is found out and then, by using the given conditions the particular solution is generated.
We are given the equation \[{\sec ^2}x = \sqrt 2 (1 - {\tan ^2}x)\]
We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]
Therefore we get
\[1 + {\tan ^2}x = \sqrt 2 (1 - {\tan ^2}x)\]
Hence on simplification we get
\[1 + {\tan ^2}x = \sqrt 2 - \sqrt 2 {\tan ^2}x\]
Hence on further simplification we get
\[{\tan ^2}x + \sqrt 2 {\tan ^2}x = \sqrt 2 - 1\]
Hence on further simplification we get
\[{\tan ^2}x(1 + \sqrt 2 ) = \sqrt 2 - 1\]
Which gives us
\[{\tan ^2}x = \dfrac{{\left( {\sqrt 2 - 1} \right)}}{{(1 + \sqrt 2 )}}\] …(1)
Now we know that
\[\tan \dfrac{\pi }{4} = 1 = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
Taking the 2nd and 3rd equality we get
\[1 = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
On cross multiplication we get
\[1 - {\tan ^2}\dfrac{\pi }{8} = 2\tan \dfrac{\pi }{8}\]
Taking all the terms on one side we get
\[{\tan ^2}\dfrac{\pi }{8} + 2\tan \dfrac{\pi }{8} - 1 = 0\]
This is a quadratic equation in \[\tan \dfrac{\pi }{8}\] .
Therefore on solving this equation by quadratic formula we get
\[\tan \dfrac{\pi }{8} = - 1 \pm \sqrt 2 \]
And hence on squaring both sides we get
\[{\tan ^2}\dfrac{\pi }{8} = \dfrac{{\left( {\sqrt 2 - 1} \right)}}{{(1 + \sqrt 2 )}}\] …(2)
Therefore by using equations (1) and (2) we get
Principle solution \[ = \dfrac{\pi }{8}\]
Therefore we get general solution \[ = n\pi \pm \dfrac{\pi }{8}\]

So, the correct answer is “Option C”.

Note: Use the trigonometric identity correctly. Try to simplify the equation as much as possible. The general quadratic equation is of the form \[a{x^2} + bx + c = 0\] where a,b and c are constants and x is a variable. Keep in mind that the particular solution and the general solution of an equation are two different things .