Question

If value of ${\left( {r\cos \theta - \sqrt 3 } \right)^2} + {\left( {r\sin \theta - 1} \right)^2} = 0$ then $\dfrac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan }} = ?$

Answer 1

We need to find the value of the given expression. We will equate both terms inside the bracket with zero to obtain the value of $\tan \theta$ from there. We will then find the value of $\sec \theta$. We will then substitute their values in the expression whose value we need to find. From there, we will get the value of the given expression.

Complete step-by-step answer:
It is given that,
${\left( {r\cos \theta - \sqrt 3 } \right)^2} + {\left( {r\sin \theta - 1} \right)^2} = 0$
This is the sum of two positive terms as the square of any term is positive. This is possible only when
$\Rightarrow$ ${\left( {r\cos \theta - \sqrt 3 } \right)^2} = 0$ and ${\left( {r\sin \theta - 1} \right)^2} = 0$.
Now solving each term individually, we get
$\Rightarrow$ ${\left( {r\cos \theta - \sqrt 3 } \right)^2} = 0$
Taking square root on both sides, we get
$\Rightarrow \sqrt {{{\left( {r\cos \theta - \sqrt 3 } \right)}^2}} = \sqrt 0 \\\\\Rightarrow r\cos \theta - \sqrt 3 = 0$
Adding $\sqrt 3$ on both sides, we get
$\Rightarrow$ $r\cos \theta - \sqrt 3 + \sqrt 3 = 0 + \sqrt 3$
$\Rightarrow$ $r\cos \theta = \sqrt 3$……………….$\left( 1 \right)$
Now, we will evaluate ${\left( {r\sin \theta - 1} \right)^2} = 0$.
Taking square root on both sides, we get
$\Rightarrow \sqrt {{{\left( {r\sin \theta - 1} \right)}^2}} = \sqrt 0 \\\\\Rightarrow r\sin \theta - 1 = 0$
Adding 1 on both sides, we get
$\Rightarrow$ $r\sin \theta - 1 + 1 = 0 + 1$
$\Rightarrow$ $r\sin \theta = 1$ ……………….$\left( 2 \right)$
We will now divide equation $\left( 1 \right)$ by equation $\left( 2 \right)$ now.
$\Rightarrow$ $\dfrac{{r\sin \theta }}{{r\cos \theta }} = \dfrac{1}{{\sqrt 3 }}$
On further simplification, we get
$\Rightarrow \tan \theta = \dfrac{1}{{\sqrt 3 }}\\\\\Rightarrow \theta = {30^0}$
We will calculate the value of $\sec \theta$now.
We know, $\sec \theta = \sqrt {1 + {{\tan }^2}\theta }$
We will put the value of $\tan \theta$
$\Rightarrow$ $\sec \theta = \sqrt {1 + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}$
Simplifying the terms, we get
$\Rightarrow$ $\sec \theta = \sqrt {1 + \dfrac{1}{3}} = \dfrac{2}{{\sqrt 3 }}$
Now we will find the value of $r$.
Squaring and adding equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we get
$\Rightarrow$ ${r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = {\left( {\sqrt 3 } \right)^2} + {1^2}$
Simplifying the equation, we get
$\Rightarrow$ ${r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = 4$
We know from trigonometric identities that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ .
Therefore we can rewrite the equation as,
${r^2} = 4$
Taking square root on both sides, we get
$\Rightarrow \sqrt {{r^2}} = \sqrt 4 \\\ \Rightarrow r = 2$
Now we will calculate the value of $\dfrac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan }}$.
Here we will put the value of $\tan \theta$, $r$ and $\sec \theta$.
$\Rightarrow$ $\dfrac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan \theta }} = \dfrac{{2.\dfrac{1}{{\sqrt 3 }} + \dfrac{2}{{\sqrt 3 }}}}{{2.\dfrac{2}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }}}}$
Simplifying the terms, we get
$\Rightarrow$ $\dfrac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan }} = \dfrac{{2 + 2}}{{4 + 1}} = \dfrac{4}{5}$
This is the required answer.

Note: Here it is important for us to remember all the identities and properties of trigonometric functions. We have also used the concept that if a and b are two numbers and it is given that ${a^2} + {b^2} = 0$ then the value of both a and b is zero. This is because ${a^2}$ is positive and ${b^2}$ is also positive, their sum is zero only when $a$ and $b$ both are zero.