Question

If $\tan A = \dfrac{1}{{\sqrt 3 }}$ then find the value of $\dfrac{{{\text{cose}}{{\text{c}}^2}A - {{\sec }^2}A}}{{{\text{cose}}{{\text{c}}^2}A + {{\sec }^2}A}}$.

Answer 1

Hint: First, we will find the value of $A$ using the given value of $\tan A$ and taking ${\tan ^{ - 1}}$ on both the sides in the given equation. Then we will substitute the value of $A$ in the given expression $\dfrac{{{\text{cose}}{{\text{c}}^2}A - {{\sec }^2}A}}{{{\text{cose}}{{\text{c}}^2}A + {{\sec }^2}A}}$ to find the required value.

Complete step-by-step answer:

It is given that$\tan A = \dfrac{1}{{\sqrt 3 }}$.

We know that the value of the tangential function at 30 degrees is $\tan 30 = \dfrac{1}{{\sqrt 3 }}$.
Using the value in the given equation, we get

$$\tan A = \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow \tan A = \tan 30 \\\$$

We know that the inverse $\tan$ is the inverse function of the trigonometric function ‘tangent’.

We also know that the property of the tangential property, ${\tan ^{ - 1}}\tan x = x$.

Taking ${\tan ^{ - 1}}$ in the above equation on each of the sides, we get

$$\Rightarrow {\tan ^{ - 1}}\tan A = {\tan ^{ - 1}}\tan 30 \\\ \Rightarrow A = 30 \\\$$

Substituting this value of $A$ in the given expression $\dfrac{{{\text{cose}}{{\text{c}}^2}A - {{\sec }^2}A}}{{{\text{cose}}{{\text{c}}^2}A + {{\sec }^2}A}}$ to find the required value, we get

$$\dfrac{{{\text{cose}}{{\text{c}}^2}30 - {{\sec }^2}30}}{{{\text{cose}}{{\text{c}}^2}30 + {{\sec }^2}30}} = \dfrac{{4 - {{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2}}}{{4 + {{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2}}} \\\ = \dfrac{{4 - \dfrac{4}{3}}}{{4 + \dfrac{4}{3}}} \\\ = \dfrac{{\dfrac{{12 - 4}}{4}}}{{\dfrac{{12 + 4}}{4}}} \\\ = \dfrac{{\dfrac{8}{4}}}{{\dfrac{{16}}{4}}} \\\ = \dfrac{8}{{16}} \\\ = \dfrac{1}{2} \\\$$

Hence, $\dfrac{{{\text{cose}}{{\text{c}}^2}A - {{\sec }^2}A}}{{{\text{cose}}{{\text{c}}^2}A + {{\sec }^2}A}} = \dfrac{1}{2}$.

Note: In this question, first of all, students should use the properties of trigonometry to make it easier to find the required value. Also, in these types of questions, we can also find the value of the angle by using the Pythagorean theorem ${h^2} = {a^2} + {b^2}$, where $h$ is the hypotenuse, $a$ is the height and $b$ is the base of the triangle, on two given sides of trigonometric function by drawing the triangle and find the third value of the triangle, only when the value cannot be calculated directly.