Question

If the value of $\sec \theta - \tan \theta = 8$
Then find the value of $\left( {\sec \theta + \tan \theta } \right)$ .

Answer 1

Hint: In this question apply the identity which is $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ and apply the basic trigonometric property which is $\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) = 1$ so use this concept to reach the solution of the question.

Complete step-by-step answer:

Given trigonometric equation is
$\sec \theta - \tan \theta = 8$……………………. (1)
Then we have to find out the value of $\left( {\sec \theta + \tan \theta } \right)$.
Now as we know that the value of $\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) = 1$………………….. (2)
And we break this equation by using $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ so, use this property to break equation (2) we have,
$ \Rightarrow \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) = \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) = 1$
Now from equation (1) we have,
$ \Rightarrow \left( {\sec \theta + \tan \theta } \right)8 = 1$
Now divide by 8 in above equation we have,
$ \Rightarrow \left( {\sec \theta + \tan \theta } \right) = \dfrac{1}{8}$
So, this is the required answer.

Note: In such types of question the key concept we have to remember is that always recall the basic trigonometric properties and the basic identity which is stated above then using these properties simplify the given equation we will get the required value of $\left( {\sec \theta + \tan \theta } \right)$ which is the required answer.