Question

If $x=\tan \theta \left| \tan \theta \right|$, $y=\cot \theta \left| \cot \theta \right|$ where $\dfrac{51\pi }{2}\le \theta \le 26\pi $, then which one of the options is true?
(a) $x+y\ge 4$
(b) $x+y\ge 2$
(c) $x+y\le -2$
(d) $x+y\le -4$

Answer 1

Hint: First of all remove the modulus sign from the terms x and y by finding the quadrant in which $\theta $ lies and considering that tangent and co – tangent is positive in the first and third quadrant. Take the sum of expressions x and y and use the conversion ${{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab$ to simplify the sum. Use the trigonometric identity $\tan \theta \times \cot \theta =1$. Use the result ${{\left( a-b \right)}^{2}}\ge 0$ and find the required inequality.

Complete step by step answer:
Here we have been provided with the expressions $x=\tan \theta \left| \tan \theta \right|$, $y=\cot \theta \left| \cot \theta \right|$ with the range of angle as $\dfrac{51\pi }{2}\le \theta \le 26\pi $. We are asked to find the correct inequality from the given options.
Now, first of all we need to remove the modulus sign by checking the sign of the functions present inside the modulus. For that we need to check the quadrant in which the angle lies. Since, the range of angle is $\dfrac{51\pi }{2}\le \theta \le 26\pi $ and we know that a full round on the circle will result in the angle $2\pi $ that means the according to the given range of $\theta $ given denotes that we took $25\dfrac{1}{2}$ rounds of the circle moved a bit ahead but did not completed the ${{26}^{th}}$ round. This clearly means we are in the ${{4}^{th}}$ quadrant.
We know that the tangent and co – tangent function is positive in first and third quadrant but negative in the second and fourth quadrant, so the two functions inside the modulus sign are negative, so we get,
$\begin{align}
  & \Rightarrow x=\tan \theta \left( -\tan \theta \right) \\
 & \Rightarrow x=-{{\tan }^{2}}\theta .............\left( i \right) \\
\end{align}$
\[\begin{align}
  & \Rightarrow y=\cot \theta \left( -\cot \theta \right) \\
 & \Rightarrow y=-{{\cot }^{2}}\theta ............\left( ii \right) \\
\end{align}\]
Adding equations (i) and (ii) we get,
$\begin{align}
  & \Rightarrow x+y=-{{\tan }^{2}}\theta -{{\cot }^{2}}\theta \\
 & \Rightarrow x+y=-\left( {{\tan }^{2}}\theta +{{\cot }^{2}}\theta \right) \\
\end{align}$
We can write the expression \[{{a}^{2}}+{{b}^{2}}\] equal to ${{\left( a-b \right)}^{2}}+2ab$, so we get,
$\Rightarrow x+y=-\left[ {{\left( \tan \theta -\cot \theta \right)}^{2}}+2\tan \theta \cot \theta \right]$
Using the identity $\tan \theta \times \cot \theta =1$ we get,
$\Rightarrow x+y=-\left[ {{\left( \tan \theta -\cot \theta \right)}^{2}}+2 \right]$
We know that ${{\left( a-b \right)}^{2}}\ge 0$ so we have,
\[\begin{align}
  & \Rightarrow {{\left( \tan \theta -\cot \theta \right)}^{2}}\ge 0 \\
 & \Rightarrow {{\left( \tan \theta -\cot \theta \right)}^{2}}+2\ge 2 \\
\end{align}\]
Multiplying both the sides with -1 and changing the direction of inequality we get,
\[\begin{align}
  & \Rightarrow -\left[ {{\left( \tan \theta -\cot \theta \right)}^{2}}+2 \right]\le -2 \\
 & \therefore x+y\le -2 \\
\end{align}\]

So, the correct answer is “Option c”.

Note: You may think that \[{{a}^{2}}+{{b}^{2}}\] can also be written as ${{\left( a+b \right)}^{2}}-2ab$ but we haven’t done do. The reason is that if we will do so then we will get the answer \[x+y\le 2\] which is correct but not exact because we have $x+y=-\left( {{\tan }^{2}}\theta +{{\cot }^{2}}\theta \right)$ which will always be negative that means it will not go beyond 0, so even if you consider this relation \[{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\] then you must also consider the relation ${{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab$ and form two sets of the inequality and take their intersection to get the answer.