Answer

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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.

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.

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