Question

How do you solve $ {\tan ^2}x + \sec x = 1 $

Answer 1

Hint: Here in this question, we have to solve the given function, the trigonometry function and it is involving the trigonometry ratios. The tangent and secant are the trigonometry ratios. By using the trigonometric identities between those two we can solve the given function.

Complete step-by-step answer:
Consider the given function $ {\tan ^2}x + \sec x = 1 $ , The tan is known as tangent and sec is known as secant are the trigonometry ratios. In the trigonometry we have three trigonometry identities and they are $ {\sin ^2}x + {\cos ^2}x = 1 $ , $ {\sec ^2}x = 1 + {\tan ^2}x $ and $ \cos e{c^2}x = 1 + {\cot ^2}x $ .
Now consider the given question
  $ {\tan ^2}x + \sec x = 1 $
By using the trigonometric identities, replace $ {\tan ^2}x $ by $ {\sec ^2}x - 1 $ , so the equation is written as
  $ \Rightarrow {\sec ^2}x - 1 + \sec x = 1 $
Take 1 from RHS to LHS and the equation is written as
  $ \Rightarrow {\sec ^2}x - 1 + \sec x - 1 = 0 $
On simplification we have
  $ \Rightarrow {\sec ^2}x + \sec x - 2 = 0 $
The above equation is in the form of a quadratic equation in terms of $ \sec x $ , by this we can find the value of $ \sec x $ .
We use formula $ \sec x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ , to find the values.
Here a=1, b=1 and c=-2. By substituting these values in the formula, we have
  $ \Rightarrow \sec x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)(-2)} }}{{2(1)}} $
On simplifying the above equation, we have
  $ \Rightarrow \sec x = \dfrac{{ - 1 \pm \sqrt {1 + 8} }}{2} $
  $ \Rightarrow \sec x = \dfrac{{ - 1 \pm \sqrt 9 }}{2} $
As we know that the square root of 9 is 3. So we get
  $ \Rightarrow \sec x = \dfrac{{ - 1 \pm 3}}{2} $
Therefore, we have $ \sec x = \dfrac{{ - 1 + 3}}{2} $ or $ \sec x = \dfrac{{ - 1 - 3}}{2} $
The values of $ \sec x $ are $ \sec x = \dfrac{2}{2} = 1 $ or $ \sec x = \dfrac{{ - 4}}{2} = - 2 $
The secant can be written in the form of cosine, so we have
  $ \dfrac{1}{{\cos x}} = 1 $ and $ \dfrac{1}{{\cos x}} = - 2 $
Therefore $ \cos x = 1 $ and $ \cos x = - \dfrac{1}{2} $
By the table of trigonometry ratios for a standard angle we can determine the value of x. therefore the value of x is $ x = 2n\pi ,\dfrac{{2\pi }}{3} $ .
Hence, we have solved the given function and determined the value of x.
So, the correct answer is “ $ x = 2n\pi ,\dfrac{{2\pi }}{3} $ ”.

Note: In trigonometry we have two main topics one is trigonometry ratios and other one is trigonometry identities. When the given function is related to the trigonometry then with the help of these two concepts, we can solve the given question further and hence we can obtain the required result.