Question

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

Answer 1

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

& \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=2 \\\ & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2=0 \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ So, $$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{aligned} & \Rightarrow \tan \theta =\tan \left( \dfrac{\pi }{4} \right) \\\ & \Rightarrow \theta ={{\tan }^{-1}}\left\\{ \tan \left( \dfrac{\pi }{4} \right) \right\\} \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \Rightarrow \tan \theta =\tan \left( \dfrac{-\pi }{4} \right) \\\ & \Rightarrow \theta ={{\tan }^{-1}}\left\\{ \tan \left( \dfrac{-\pi }{4} \right) \right\\} \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\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.