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

Hint: 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.