Question

Solve: \[\sec x - 1 = (\sqrt 2 - 1)\tan x,x \ne (2n - 1)\dfrac{\pi }{2},n \in Z\]

Answer 1

Hint: We start solving this equation by squaring both sides of the equation. Then convert the complete equation into a single trigonometric function using the identity \[1 + {\tan ^2}x = {\sec ^2}x\]. Make the quadratic equation so formed in simple form and calculate the roots of the quadratic equation. From the values obtained as roots, find the value of ‘x’.
* A quadratic equation is an equation with the highest power of the variable as 2. Roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Step-By-Step answer:
We are given an equation \[\sec x - 1 = (\sqrt 2 - 1)\tan x\] … (1)
Squaring both sides of the equation
\[ \Rightarrow {\left( {\sec x - 1} \right)^2} = {\left( {(\sqrt 2 - 1)\tan x} \right)^2}\]
Use identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]to open square in left hand side of the equation
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = {\left( {\sqrt 2 - 1} \right)^2}{\tan ^2}x\]
Use identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]to open square in right hand side of the equation
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = \left\{ {{{(\sqrt 2 )}^2} + 1 - 2\sqrt 2 } \right\}{\tan ^2}x\]
Cancel square root by square power in RHS of the equation
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = \left\{ {2 + 1 - 2\sqrt 2 } \right\}{\tan ^2}x\]
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = \left( {3 - 2\sqrt 2 } \right){\tan ^2}x\]
Since we know \[1 + {\tan ^2}x = {\sec ^2}x\], then we can substitute the value of \[{\tan ^2}x = {\sec ^2}x - 1\] in RHS
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = \left( {3 - 2\sqrt 2 } \right)\left( {{{\sec }^2}x - 1} \right)\]
Multiply the terms in RHS of the equation
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x = 3{\sec ^2}x - 3 - 2\sqrt 2 {\sec ^2}x + 2\sqrt 2 \]
Shift all values to left hand side of the equation
\[ \Rightarrow {\sec ^2}x + 1 - 2\sec x - 3{\sec ^2}x + 3 + 2\sqrt 2 {\sec ^2}x - 2\sqrt 2 = 0\]
Combine all terms with same variables
\[ \Rightarrow {\sec ^2}x(1 - 3 + 2\sqrt 2 ) - (2\sec x) + (3 + 1 - 2\sqrt 2 ) = 0\]
\[ \Rightarrow {\sec ^2}x( - 2 + 2\sqrt 2 ) - (2\sec x) + (4 - 2\sqrt 2 ) = 0\]
Take 2 common from all the terms
\[ \Rightarrow 2\left\{ {{{\sec }^2}x( - 1 + \sqrt 2 ) - (\sec x) + (2 - \sqrt 2 )} \right\} = 0\]
Divide both sides of the equation by 2
\[ \Rightarrow {\sec ^2}x( - 1 + \sqrt 2 ) - \sec x + (2 - \sqrt 2 ) = 0\] … (2)
This forms a quadratic equation in secant
Comparing equation (2) with general quadratic equation \[a{x^2} + bx + c = 0\],
\[a = \sqrt 2 - 1;b = - 1;c = 2 - \sqrt 2 \]
Then the roots of equation (2) are given by \[\sec x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute values of ‘a’, ‘b’ and ‘c’ in the formula of roots
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {1 - 4\left\{ {\sqrt 2 \times 2 - \sqrt 2 \times \sqrt 2 - 1 \times 2 - 1 \times - \sqrt 2 } \right\}} }}{{2(\sqrt 2 - 1)}}\]
Solve the value under the square root
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {1 - 4\left\{ {2\sqrt 2 - 2 - 2 + \sqrt 2 } \right\}} }}{{2(\sqrt 2 - 1)}}\]
Add terms in numerator
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {1 - 4\left\{ {3\sqrt 2 - 4} \right\}} }}{{2(\sqrt 2 - 1)}}\]
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {1 - 12\sqrt 2 + 16} }}{{2(\sqrt 2 - 1)}}\]
Add terms in numerator
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {17 - 12\sqrt 2 } }}{{2(\sqrt 2 - 1)}}\] … (3)
Now we know \[{\left( {3 - 2\sqrt 2 } \right)^2} = {(3)^2} + {(2\sqrt 2 )^2} - 2 \times 3 \times 2\sqrt 2 \]
i.e. \[{\left( {3 - 2\sqrt 2 } \right)^2} = 9 + 8 - 12\sqrt 2 \]
i.e. \[{\left( {3 - 2\sqrt 2 } \right)^2} = 17 - 12\sqrt 2 \]
Substitute the value of \[{\left( {3 - 2\sqrt 2 } \right)^2} = 17 - 12\sqrt 2 \] in equation (3)
\[ \Rightarrow \sec x = \dfrac{{1 \pm \sqrt {{{(3 - 2\sqrt 2 )}^2}} }}{{2(\sqrt 2 - 1)}}\]
Cancel square root by square power in numerator
\[ \Rightarrow \sec x = \dfrac{{1 \pm \left( {3 - 2\sqrt 2 } \right)}}{{2(\sqrt 2 - 1)}}\]
From here we get two roots
\[ \Rightarrow \sec x = \dfrac{{1 + \left( {3 - 2\sqrt 2 } \right)}}{{2(\sqrt 2 - 1)}}\] and \[\sec x = \dfrac{{1 - \left( {3 - 2\sqrt 2 } \right)}}{{2(\sqrt 2 - 1)}}\]
\[ \Rightarrow \sec x = \dfrac{{4 - 2\sqrt 2 }}{{2(\sqrt 2 - 1)}}\]and \[\sec x = \dfrac{{ - 2 + 2\sqrt 2 }}{{2(\sqrt 2 - 1)}}\]
Cancel same factors from numerator and denominator i.e. 2
\[ \Rightarrow \sec x = \dfrac{{2 - \sqrt 2 }}{{(\sqrt 2 - 1)}}\]and \[\sec x = \dfrac{{ - 1 + \sqrt 2 }}{{(\sqrt 2 - 1)}}\]
Cancel same factors from numerator and denominator i.e.2
\[ \Rightarrow \sec x = \dfrac{{2 - \sqrt 2 }}{{(\sqrt 2 - 1)}}\]and \[\sec x = 1\]
Rationalize the root in left side
\[ \Rightarrow \sec x = \dfrac{{2 - \sqrt 2 }}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}}\] and \[\sec x = 1\]
Use identity \[(a - b)(a + b) = {a^2} - {b^2}\] in denominator
\[ \Rightarrow \sec x = \dfrac{{2\sqrt 2 - 2 - 2 + \sqrt 2 }}{{{{\left( {\sqrt 2 } \right)}^2} - 1}}\] and \[\sec x = 1\]
\[ \Rightarrow \sec x = \dfrac{{3\sqrt 2 - 4}}{{2 - 1}}\] and \[\sec x = 1\]
\[ \Rightarrow \sec x = 3\sqrt 2 - 4\] and \[\sec x = 1\]
Since we know \[3 < 4\], then the value \[\sec x = 3\sqrt 2 - 4 < 1\]
So we ignore the first root
Root of quadratic equation is \[\sec x = 1\]
Now we know \[\cos {0^ \circ } = 1\], then \[\sec {0^ \circ } = 1\] as cosine and secant are reciprocal of each other.
\[ \Rightarrow x = 2n\pi + {0^ \circ }\]
\[ \Rightarrow x = 2n\pi ,n \in Z\]

\[\therefore \]The values of x are \[x = 2n\pi ,n \in Z\]

Note: Many students make the mistake of solving the equation by converting all terms in sine and cosine form. In this case we don’t proceed in this order as we will have to do these extra steps and then form an equation. Also, keep in mind the roots cannot be negative because values of secant has to be positive.