Question

If ${\tan ^2}\theta + \sec \theta = 5$ , find $\cos \theta $ .

Answer 1

Hint: Firstly, use ${\tan ^2}\theta = {\sec ^2}\theta - 1$ . After that, a quadratic equation will be formed. By solving that quadratic equation, using the method of splitting the middle term, we will get max. two values of $\sec \theta $ . And by taking the reciprocal of $\sec \theta $ , we will get the value of $\cos \theta $ .

Complete step-by-step answer:
We have
 ${\tan ^2}\theta + \sec \theta = 5$
We know that, ${\tan ^2}\theta = {\sec ^2}\theta - 1$
 $
  \therefore {\sec ^2}\theta - 1 + \sec \theta - 5 = 0 \\
  \therefore {\sec ^2}\theta + \sec \theta - 6 = 0 \\
 $
Now, on splitting the middle term $\sec \theta $ as $3\sec \theta - 2\sec \theta $ we get,
 $
  {\sec ^2}\theta + 3\sec \theta - 2\sec \theta - 6 = 0 \\
  \therefore \sec \theta \left( {\sec \theta + 3} \right) - 2\left( {\sec \theta + 3} \right) = 0 \\
  \therefore \left( {\sec \theta + 3} \right)\left( {\sec \theta - 2} \right) = 0 \\
 $
 $\therefore \sec \theta + 3 = 0$ or $\sec \theta - 2 = 0$
 $\therefore \sec \theta = - 3$ or $\sec \theta = 2$
 $\therefore \dfrac{1}{{\cos \theta }} = - 3$ or $\dfrac{1}{{\cos \theta }} = 2$
 $\therefore \cos \theta = \dfrac{{ - 1}}{3}$ or $\cos \theta = \dfrac{1}{2}$
Thus, $\cos \theta = \dfrac{{ - 1}}{3}$ or $\cos \theta = \dfrac{1}{2}$ .

Note: Alternate Method:
We have
 ${\tan ^2}\theta + \sec \theta = 5$
We know that, ${\tan ^2}\theta = {\sec ^2}\theta - 1$
 $
  \therefore {\sec ^2}\theta - 1 + \sec \theta - 5 = 0 \\
  \therefore {\sec ^2}\theta + \sec \theta - 6 = 0 \\
 $
Now, from the above equation, \[a = 1\] , \[b = 1\] and \[c = - 6\] .
 $\therefore \Delta = \sqrt {{b^2} - 4ac} $
         $
   = \sqrt {{1^2} - 4\left( 1 \right)\left( { - 6} \right)} \\
   = \sqrt {1 + 24} \\
   = \sqrt {25} \\
   = 5 \\
 $
So, the roots of the equation are $\sec \theta = \dfrac{{ - b \pm \Delta }}{{2a}}$ .
 $\therefore \sec \theta = \dfrac{{ - b + \Delta }}{{2a}}$ or $\sec \theta = \dfrac{{ - b - \Delta }}{{2a}}$
 $\therefore \sec \theta = \dfrac{{ - 1 + 5}}{2}$ or $\sec \theta = \dfrac{{ - 1 - 5}}{2}$
 $\therefore \sec \theta = \dfrac{4}{2}$ or $\sec \theta = \dfrac{{ - 6}}{2}$
 $\therefore \sec \theta = 2$ or $\sec \theta = - 3$
 $\therefore \dfrac{1}{{\cos \theta }} = 2$ or $\dfrac{1}{{\cos \theta }} = - 3$
 $\therefore \cos \theta = \dfrac{1}{2}$ or $\cos \theta = \dfrac{{ - 1}}{3}$
Thus, $\cos \theta = \dfrac{{ - 1}}{3}$ or $\cos \theta = \dfrac{1}{2}$ .