Question
Answers

Eliminate $\theta $, if $r=y\csc \theta \text{ and }x=y\cot \theta $.

Answer Verified Verified
Hint: Try to eliminate $\theta $ by using trigonometric identity $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $ or ${{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1$, from the given relations of $x,y,r\ and\ \theta $.

Complete step-by-step answer:
We have relation given as
$\begin{align}
  & r=y\csc \theta .............\left( 1 \right) \\
 & x=y\cot \theta ..............\left( 2 \right) \\
\end{align}$
We can get the value of $\csc \theta $ from equation (1) by transferring y to the other side.
Therefore, equation (1), can be written as
$\csc \theta =\dfrac{r}{y}...............\left( 3 \right)$
Now, value of $\cot \theta $ can be given by equation (2) by transferring y to other side, we get
$\cot \theta =\dfrac{x}{y}...................\left( 4 \right)$
Now, we know trigonometric identities related to trigonometric functions $\cot \theta $and $\csc \theta $ as;
${{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1..............\left( 5 \right)$
Now, we can put the value of $\cot \theta $and $\csc \theta $to the equation (5) from the equation (3) and (4).
Value of ${{\csc }^{2}}\theta $from equation (3) is $\dfrac{{{r}^{2}}}{{{y}^{2}}}$
Therefore, ${{\csc }^{2}}\theta =\dfrac{{{r}^{2}}}{{{y}^{2}}}$
Value of ${{\cot }^{2}}\theta $ from equation (4) is \[\dfrac{{{x}^{2}}}{{{y}^{2}}}\] .
Hence we get
\[{{\cot }^{2}}\theta =\dfrac{{{x}^{2}}}{{{y}^{2}}}\]
Now, we have ${{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1$ from the equation (5), hence by putting values of ${{\csc }^{2}}\theta \text{ and }{{\cot }^{2}}\theta $ as calculated above be
$\dfrac{{{r}^{2}}}{{{y}^{2}}}-\dfrac{{{x}^{2}}}{{{y}^{2}}}=1$
Taking ${{y}^{2}}$ as L.C.M of LHS part, we get;
$\dfrac{{{r}^{2}}-{{x}^{2}}}{{{y}^{2}}}=\dfrac{1}{1}$
On cross multiplying the terms, we get
\[{{r}^{2}}-{{x}^{2}}={{y}^{2}}\]
Transferring x to other side, we get
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]
Hence, by eliminating $\theta $ from $r=y\csc \theta ,x=y\cot \theta $, We get;
\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\].

Note: Another approach for this question would be that we can convert the above relation to sin and cos.
We have $r=y\csc \theta ,\text{ as }\csc \theta =\dfrac{1}{\sin \theta }$
Hence, we get $r=\dfrac{y}{\sin \theta }$
Now, from the second relation i.e. $x=y\cot \theta $we know$\cot \theta =\dfrac{\cos \theta }{\sin \theta }$;
Hence, we can write it as
$\dfrac{\cos \theta }{\sin \theta }=\dfrac{x}{y}$
One can go wrong while using the relation${{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1$. One can confuse the relation ${{\cot }^{2}}\theta -{{\csc }^{2}}\theta =1$ which is wrong and will give incorrect answers.
Bookmark added to your notes.
View Notes
×