Question

${{\cot }^{-1}}\left( \dfrac{y}{\sqrt{1-{{x}^{2}}-{{y}^{2}}}} \right)=2{{\tan }^{-1}}\left( \sqrt{\dfrac{3-4{{x}^{2}}}{4{{x}^{2}}}} \right)-{{\tan }^{-1}}\left( \sqrt{\dfrac{3-4{{x}^{2}}}{{{x}^{2}}}} \right)$
Expression of I as a rational integral equation in x and y is
(a) $27{{x}^{2}}={{y}^{2}}{{(9-8{{y}^{2}})}^{2}}$
(b) $27{{y}^{2}}={{x}^{2}}{{(9-8{{x}^{2}})}^{2}}$
(c) $27{{x}^{4}}-9{{x}^{2}}+8{{y}^{2}}=0$
(d) $27{{y}^{4}}-9{{y}^{2}}+8{{x}^{2}}=0$

Answer 1

Hint:We will first simplify left hand side of the equation by using inverse trigonometric formulas $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$. Then we will equate it to the right hand side of the expression by converting cot in terms of tan.

Complete step-by-step answer:
Before proceeding with the question let us know some of the inverse trigonometric formulas. We know that $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)........(1)$.
We also know that ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)........(2)$.
First assuming $t=\left( \sqrt{\dfrac{3-4{{x}^{2}}}{{{x}^{2}}}} \right)$. We will first simplify the first term of the right hand side of the expression by using equation (1) and substituting t and we get,
$\Rightarrow 2{{\tan }^{-1}}\left( \sqrt{\dfrac{3-4{{x}^{2}}}{4{{x}^{2}}}} \right)=2{{\tan }^{-1}}\dfrac{t}{2}={{\tan }^{-1}}\dfrac{2\times \dfrac{t}{2}}{1-\dfrac{{{t}^{2}}}{4}}.........(3)$
Now rearranging equation (3) by taking LCM and cancelling similar terms we get,
$\Rightarrow {{\tan }^{-1}}\dfrac{4t}{4-{{t}^{2}}}.........(4)$
Now right hand side of the expression is,
$\Rightarrow 2{{\tan }^{-1}}\left( \sqrt{\dfrac{3-4{{x}^{2}}}{4{{x}^{2}}}} \right)-{{\tan }^{-1}}\left( \sqrt{\dfrac{3-4{{x}^{2}}}{{{x}^{2}}}} \right).....(5)$
Substituting for the first term in equation (5) from equation (4) we get,
$\Rightarrow {{\tan }^{-1}}\dfrac{4t}{4-{{t}^{2}}}-{{\tan }^{-1}}t.....(6)$
Now using equation (2) in equation (6) we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{4t}{4-{{t}^{2}}}-t}{1+\dfrac{4t}{4-{{t}^{2}}}\times t} \right)={{\tan }^{-1}}\left( \dfrac{4t-4t+{{t}^{3}}}{4-{{t}^{2}}+4{{t}^{2}}} \right).....(7)$
Now cancelling similar terms in equation (7) we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{{{t}^{3}}}{4+3{{t}^{2}}} \right).....(8)$
Now converting the left hand side of the expression in terms of tan inverse we get,
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{y}{\sqrt{1-{{x}^{2}}-{{y}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}-{{y}^{2}}}}{y} \right)........(9)$
Now we equate equation (8) and equation (9),
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}-{{y}^{2}}}}{y} \right)={{\tan }^{-1}}\left( \dfrac{{{t}^{3}}}{4+3{{t}^{2}}} \right).....(10)$
So from equation (10) tan inverse gets cancelled on both sides and we get,
$\Rightarrow \left( \sqrt{\dfrac{1-{{x}^{2}}-{{y}^{2}}}{{{y}^{2}}}} \right)=\left( \dfrac{{{t}^{3}}}{4+3{{t}^{2}}} \right).....(11)$
Squaring both sides of equation (11) and rearranging equation (11) we get,

& \Rightarrow {{\left( \sqrt{\dfrac{1-{{x}^{2}}-{{y}^{2}}}{{{y}^{2}}}} \right)}^{2}}={{\left( \dfrac{{{t}^{3}}}{4+3{{t}^{2}}} \right)}^{2}} \\\ & \Rightarrow \dfrac{1-{{x}^{2}}-{{y}^{2}}}{{{y}^{2}}}=\dfrac{{{t}^{6}}}{16+9{{t}^{4}}+24{{t}^{2}}}.....(12) \\\ \end{aligned}$$ Now separately dividing the terms by denominator in left hand side of equation (12) we get, $$\Rightarrow \dfrac{1-{{x}^{2}}}{{{y}^{2}}}-1=\dfrac{{{t}^{6}}}{16+9{{t}^{4}}+24{{t}^{2}}}.....(13)$$ And again rearranging in equation (13) by taking LCM we get, $$\begin{aligned} & \Rightarrow \dfrac{1-{{x}^{2}}}{{{y}^{2}}}=\dfrac{{{t}^{6}}}{16+9{{t}^{4}}+24{{t}^{2}}}+1 \\\ & \Rightarrow \dfrac{1-{{x}^{2}}}{{{y}^{2}}}=\dfrac{{{t}^{6}}+16+9{{t}^{4}}+24{{t}^{2}}}{16+9{{t}^{4}}+24{{t}^{2}}}.........(14) \\\ \end{aligned}$$ Now cross multiplying the terms in equation (14) we get, $$\Rightarrow {{y}^{2}}=\dfrac{9{{t}^{4}}+24{{t}^{2}}+16}{{{t}^{6}}+9{{t}^{4}}+24{{t}^{2}}+16}(1-{{x}^{2}}).........(15)$$ Now we know $$t=\left( \sqrt{\dfrac{3-4{{x}^{2}}}{{{x}^{2}}}} \right)$$. So squaring both sides we get $${{t}^{2}}=\dfrac{3-4{{x}^{2}}}{{{x}^{2}}}=\dfrac{3}{{{x}^{2}}}-4$$ and then after rearranging it we get $${{t}^{2}}+4=\dfrac{3}{{{x}^{2}}}$$and then converting the terms in equation (15) in $${{t}^{2}}+4$$ form and then substituting it in equation (15) we get, $$\Rightarrow {{y}^{2}}=\dfrac{{{({{t}^{2}}+4)}^{2}}+8{{t}^{4}}+16{{t}^{2}}}{{{t}^{2}}({{t}^{4}}+8{{t}^{2}}+16)+({{t}^{4}}+8{{t}^{2}}+16)}(1-{{x}^{2}}).........(16)$$ Now simplifying equation (16) by rearranging the terms we get, $$\begin{aligned} & \Rightarrow {{y}^{2}}=\dfrac{{{({{t}^{2}}+4)}^{2}}+8{{t}^{4}}+16{{t}^{2}}}{({{t}^{2}}+1)({{t}^{4}}+8{{t}^{2}}+16)}(1-{{x}^{2}}) \\\ & \Rightarrow {{y}^{2}}=\dfrac{{{({{t}^{2}}+4)}^{2}}+8{{t}^{2}}({{t}^{2}}+2)}{({{t}^{2}}+1){{({{t}^{2}}+4)}^{2}}}(1-{{x}^{2}})........(17) \\\ \end{aligned}$$ Now substituting all the values of $${{t}^{2}}$$ in equation (17) we get, $$\Rightarrow {{y}^{2}}=\dfrac{{{\left( \dfrac{3}{{{x}^{2}}} \right)}^{2}}+8\left( \dfrac{3}{{{x}^{2}}}-4 \right)\left( \dfrac{3}{{{x}^{2}}}-2 \right)}{\left( \dfrac{3}{{{x}^{2}}}-3 \right){{\left( \dfrac{3}{{{x}^{2}}} \right)}^{2}}}(1-{{x}^{2}})........(18)$$ Now rearranging again in equation (18) we get, $$\Rightarrow {{y}^{2}}=\dfrac{\left( \dfrac{9}{{{x}^{4}}} \right)+8\left( \dfrac{3}{{{x}^{2}}}-4 \right)\left( \dfrac{3}{{{x}^{2}}}-2 \right)}{3\left( \dfrac{1-{{x}^{2}}}{{{x}^{2}}} \right)\left( \dfrac{9}{{{x}^{4}}} \right)}(1-{{x}^{2}})........(19)$$ Now cancelling all the similar terms and rearranging in equation (19) we get, $$\Rightarrow {{y}^{2}}=\dfrac{{{x}^{6}}\left( \dfrac{9}{{{x}^{4}}} \right)+8{{x}^{6}}\left( \dfrac{3}{{{x}^{2}}}-4 \right)\left( \dfrac{3}{{{x}^{2}}}-2 \right)}{27}........(20)$$ Now cross multiplying in equation (20) we get, $$\begin{aligned} & \Rightarrow 27{{y}^{2}}=9{{x}^{2}}+8{{x}^{6}}\left( \dfrac{9}{{{x}^{4}}}-\dfrac{6}{{{x}^{2}}}-\dfrac{12}{{{x}^{2}}}+8 \right) \\\ & \Rightarrow 27{{y}^{2}}=81{{x}^{2}}-144{{x}^{4}}+64{{x}^{6}}.......(21) \\\ \end{aligned}$$ Now converting right hand side of equation (21) in terms of square we get, $$\Rightarrow 27{{y}^{2}}={{x}^{2}}{{(9-8{{x}^{2}})}^{2}}$$ Hence rational integral in x and y is option (b). Note: Here we have to remember the trigonometric formulas because these formulas are very important to solve these types of questions. We can commit a mistake in solving equation (3) and equation (6) in a hurry so we need to be careful with the steps.