Question

If $\csc \theta -\cot \theta =k$, then find the value of $\tan \dfrac{\theta }{2}$.

Answer 1

Hint: Use the fact that ${{\csc }^{2}}x-{{\cot }^{2}}x=1$. Hence prove that $\left( \csc x-\cot x \right)\left( \csc x+\cot x \right)=1$ and hence find the value of $\csc \theta +\cot \theta $. Hence find the value of $\cot \theta $. Using $\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}$, form a quadratic equation in $\tan \dfrac{\theta }{2}$ and hence find the value of $\tan \dfrac{\theta }{2}$. Alternatively, express $\csc \theta -\cot \theta $ in terms of $\sin \theta $ and $\cos \theta $. Use $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.

Complete step-by-step answer:
We know that ${{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1$
Using $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, we get
$\left( \csc \theta -\cot \theta \right)\left( \csc \theta +\cot \theta \right)=1$
Now, we know that $\csc \theta -\cot \theta =k\text{ (i)}$
Hence, we have
$k\left( \csc \theta +\cot \theta \right)=1$
Dividing by k on both sides, we get
$\csc \theta +\cot \theta =\dfrac{1}{k}\text{ (ii)}$
Subtracting equation (ii), from equation (i), we get
$2\cot \theta =\dfrac{1}{k}-k$
Hence we have $2\cot \theta =\dfrac{1-{{k}^{2}}}{k}$
Taking reciprocals on both sides, we get
$\dfrac{\tan \theta }{2}=\dfrac{k}{1-{{k}^{2}}}$
Multiplying by 2 on both sides, we get
$\tan \theta =\dfrac{2k}{1-{{k}^{2}}}$
Now, we know that $\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}$
Hence, we have
$\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}=\dfrac{2k}{1-{{k}^{2}}}$
Let $\tan \dfrac{\theta }{2}=x$
Hence, we have $\dfrac{2x}{1-{{x}^{2}}}=\dfrac{2k}{1-{{k}^{2}}}$
Cross multiplying, we get
$\begin{align}
  & 2x\left( 1-{{k}^{2}} \right)=2k\left( 1-{{x}^{2}} \right) \\
 & \Rightarrow 2x\left( 1-{{k}^{2}} \right)=2k-2k{{x}^{2}} \\
\end{align}$
Dividing by 2 on both sides, we get
$\left( 1-{{k}^{2}} \right)x=k-k{{x}^{2}}$
Adding $k{{x}^{2}}$ on both sides, we get
$k{{x}^{2}}+\left( 1-{{k}^{2}} \right)x=k$
Subtracting k from both sides, we get
$k{{x}^{2}}+\left( 1-{{k}^{2}} \right)x-k=0$
Now, we know that the roots of the quadratic expression $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here $a=k,b=1-{{k}^{2}},c=-k$
Hence , we have $x=\dfrac{{{k}^{2}}-1\pm \sqrt{{{\left( 1-{{k}^{2}} \right)}^{2}}+4{{k}^{2}}}}{2k}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. Hence, we have
$x=\dfrac{{{k}^{2}}-1\pm \sqrt{1-2{{k}^{2}}+{{k}^{4}}+4{{k}^{2}}}}{2k}=\dfrac{{{k}^{2}}-1\pm \sqrt{1+2{{k}^{2}}+{{k}^{4}}}}{2k}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Hence, we have
$x=\dfrac{{{k}^{2}}-1\pm \sqrt{{{\left( {{k}^{2}}+1 \right)}^{2}}}}{2k}=\dfrac{{{k}^{2}}-1+{{k}^{2}}+1}{2k},\dfrac{{{k}^{2}}-1-{{k}^{2}}-1}{2k}=k,\dfrac{-1}{k}$
If $\tan \dfrac{\theta }{2}=\dfrac{-1}{k}$, we have $\sin \dfrac{\theta }{2}=\pm \dfrac{1}{\sqrt{1+{{k}^{2}}}},\cos \dfrac{\theta }{2}=\mp \dfrac{k}{\sqrt{1+{{k}^{2}}}}$
Hence $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}=-\dfrac{2k}{1+{{k}^{2}}}\Rightarrow \csc \theta =-\dfrac{1+{{k}^{2}}}{2k}$
Hence $\csc \theta -\cot \theta =-\dfrac{1+{{k}^{2}}}{2k}-\dfrac{2k}{1-{{k}^{2}}}=-\left( \dfrac{1-{{k}^{4}}+4{{k}^{2}}}{2k\left( 1-{{k}^{2}} \right)} \right)\ne \dfrac{1}{k}$
Hence $\tan \dfrac{\theta }{2}=\dfrac{-1}{k}$ is rejected.
Hence $\tan \dfrac{\theta }{2}=k$

Note: Alternative solution: Best method:
We have $\csc \theta -\cos \theta =\dfrac{1-\cos \theta }{\sin \theta }$
We know that $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.
Hence $\csc \theta -\cot \theta =\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}=\tan \dfrac{\theta }{2}$
Hence, we have
$\tan \dfrac{\theta }{2}=k$