Question

If the value of \[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2\] then the value of \[\theta \] is

Answer 1

Hint: First of all, assume that \[\tan \theta =x\] . Use the relation \[\tan \theta =\dfrac{1}{\cot \theta }\] and get the value of
\[{{\cot }^{2}}\theta \] in terms of \[x\] . Now, replace \[\tan \theta \] and \[\cot \theta \] in terms of \[x\] in the equation \[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2\] . Simplify it by using the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2\times a\times b\] and get the value of \[x\] . Now, use \[\tan \left( \dfrac{\pi }{4} \right)=1\] , \[\tan \left( \dfrac{5\pi }{4} \right)=1\] , \[\tan \left( \dfrac{-\pi }{4} \right)=-1\] , and \[\tan \left( \dfrac{3\pi }{4} \right)=-1\] . Then, find the general solution of \[\theta \] for \[\tan \theta =\pm 1\].

Complete step-by-step solution:
According to the question, we are given that,
\[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2\] ……………………………………………(1)
First of all, let us assume that \[\tan \theta =x\] …………………………………….(2)
We know the relation between \[\tan \theta \] and \[\cot \theta \] , \[\tan \theta =\dfrac{1}{\cot \theta }\] ………………………….(3)
On squaring the LHS and RHS of equation (3), we get
 \[\Rightarrow {{\left( \tan \theta \right)}^{2}}={{\left( \dfrac{1}{\cot \theta } \right)}^{2}}\]
\[\Rightarrow {{\tan }^{2}}\theta =\dfrac{1}{{{\cot }^{2}}\theta }\] …………………………………………….(4)
Now, using equation (2) and on substituting \[{{\tan }^{2}}\theta \] by \[x\] in equation (4), we get
\[\Rightarrow {{x}^{2}}=\dfrac{1}{{{\cot }^{2}}\theta }\]
\[\Rightarrow {{\cot }^{2}}\theta =\dfrac{1}{{{x}^{2}}}\] ……………………………………..(5)
From equation (1), equation (2), and equation (5), we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=2 \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2=0 \\
\end{align}\]
\[\Rightarrow {{\left( x \right)}^{2}}+\dfrac{1}{{{\left( x \right)}^{2}}}-2\times x\times \dfrac{1}{x}=0\] …………………………………………(6)
We know the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2\times a\times b\] …………………………………………(7)
In equation (7), on replacing \[a\] by \[x\] and \[b\] by \[\dfrac{1}{x}\] , we get
\[{{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2\times x\times \dfrac{1}{x}\] ……………………………………………..(8)
On comparing equation (6) and equation (8), we get
\[\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}=0\] ………………………………………(9)
Now, applying square root in both LHS and RHs of equation (9), we get
\[\begin{align}
  & \Rightarrow \sqrt{{{\left( x-\dfrac{1}{x} \right)}^{2}}}=\sqrt{0} \\
 & \Rightarrow \left( x-\dfrac{1}{x} \right)=0 \\
 & \Rightarrow x=\dfrac{1}{x} \\
 & \Rightarrow {{x}^{2}}=1 \\
 & \Rightarrow x=\sqrt{1} \\
 & \Rightarrow x=\pm 1 \\
\end{align}\]
So, \[x=1\] or \[x=-1\] …………………………………..(10)
From equation (2) and equation (10), we can say that
\[\tan \theta =1\] ………………………………………(11)
\[\tan \theta =-1\] …………………………………………(12)
We know that \[\tan \left( \dfrac{\pi }{4} \right)=1\] …………………………………(13)
Now, from equation (11) and equation (13), we get
\[\begin{align}
  & \Rightarrow \tan \theta =\tan \left( \dfrac{\pi }{4} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left\{ \tan \left( \dfrac{\pi }{4} \right) \right\} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{\pi }{4}\] ……………………………………(14)
We also know that \[\tan \left( \dfrac{5\pi }{4} \right)=1\] ………………………………………(15)
From equation (11) and equation (15), we get
\[\begin{align}
  & \Rightarrow \tan \theta =\tan \left( \dfrac{5\pi }{4} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left\{ \tan \left( \dfrac{5\pi }{4} \right) \right\} \\
 & \Rightarrow \theta =\dfrac{5\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\pi +\dfrac{\pi }{4}\] ……………………………………(16)
Now, from equation (14) and equation (16), we can say that the general solution of \[\tan \theta =1\] is \[\theta =n\pi +\dfrac{\pi }{4}\] ………………………………………(17)
We also know that \[\tan \left( \dfrac{-\pi }{4} \right)=-1\] …………………………………(18)
Now, from equation (12) and equation (18), we get
\[\begin{align}
  & \Rightarrow \tan \theta =\tan \left( \dfrac{-\pi }{4} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left\{ \tan \left( \dfrac{-\pi }{4} \right) \right\} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{-\pi }{4}\] ……………………………………(19)
We also know that \[\tan \left( \dfrac{3\pi }{4} \right)=-1\] ………………………………………(20)
From equation (12) and equation (20), we get
\[\begin{align}
  & \Rightarrow \tan \theta =\tan \left( \dfrac{3\pi }{4} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left\{ \tan \left( \dfrac{3\pi }{4} \right) \right\} \\
 & \Rightarrow \theta =\dfrac{3\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\pi -\dfrac{\pi }{4}\] ……………………………………(21)
Now, from equation (19) and equation (21), we can say that the general solution of \[\tan \theta =-1\] is \[\theta =n\pi -\dfrac{\pi }{4}\] ………………………………………(22)
On combining equation (17) and equation (22), we get
\[\theta =n\pi \pm \dfrac{\pi }{4}\] …………………………………………..(23)
Hence, the value of \[\theta \] is \[\left( n\pi \pm \dfrac{\pi }{4} \right)\].

Note: To solve this question easily and within the given time constraint one must remember that the general solution of \[\tan \theta =\pm 1\] is \[\left( n\pi \pm \dfrac{\pi }{4} \right)\] . If somehow one forgets this then, he must know the procedure to find out the general solution.