# Find the value of ## (ax){^\frac{2}{3}} + (by){^\frac{2}{3}}##

sahilmm15
Homework Statement:
Given, if ##{\frac{ax}{cos\theta}} + {\frac{by}{sin\theta}} = a^2-b^2## and ##{\frac{axsin\theta}{cos^2\theta}} - {\frac{bycos\theta}{sin^2\theta}} = 0## find the value of ## (ax){^\frac{2}{3}} + (by){^\frac{2}{3}}##
Relevant Equations:
N/A, can be done through general algebraic techniques.
I was trying this problem from some 2 hours but unable to crack it. I cannot proceed after few lines i.e$${\frac{axsin\theta}{cos^2\theta}} - {\frac{bycos\theta}{sin^2\theta}} = 0$$ $${\frac{axsin\theta}{cos^2\theta}} = {\frac{bycos\theta}{sin^2\theta}}$$ $${\frac{ax}{by}} ={\frac{cos^3\theta}{sin^3\theta}}$$ at last $$cot\theta = {\frac {ax^\frac{1}{3}} {by^\frac{1}{3}}}$$ This is all what I am able to do.

Gold Member
$$by=ax\ tan^3\theta$$
You can delete ##by## by it.

sahilmm15
$$by=ax\ tan^3\theta$$
You can delete ##by## by it.
It is looking horrible.

sahilmm15
It is looking horrible.
Too lengthy I guess, also note I didn't found use of the 1st equation, so maybe we can find something there.

sahilmm15
This problem is fairly difficult as it was asked in JEE Advanced.

archaic
we mustn't have ##a## and ##b## in the result?

sahilmm15
we mustn't have ##a## and ##b## in the result?
We can have ##a## and ##b##.

Gold Member
Try
$$\frac{ax}{cos\theta}+\frac{by}{sin\theta}=\frac{ax}{cos\theta}[1+\frac{\sin^2\theta}{cos^2\theta}]=...=a^2-b^2$$
You will get ax as function of a, b and ##\theta##. Then
$$(ax)^{2/3}+(by)^{2/3}=(ax)^{2/3}[ 1 + tan^2\theta]=...$$

Delta2
sahilmm15
Try
$$\frac{ax}{cos\theta}+\frac{by}{sin\theta}=\frac{ax}{cos\theta}[1+\frac{\sin^2\theta}{cos^2\theta}]=...=a^2-b^2$$
You will get ax by function of a, b and ##\theta##. Then
$$(ax)^{2/3}+(by)^{2/3}=(ax)^{2/3}[ 1 + tan^2\theta]=...$$
I am close I guess, need to convert into latex which is another nightmare for me because am just a beginner.

archaic
We can have ##a## and ##b##.
ok, then, perhaps, if you put ##A=ax## and ##B=by##, the system will look cleaner and clearer.
$$(1):\,\left(\frac{1}{\cos\theta}\right)A+\left(\frac{1}{\sin\theta}\right)B=a^2-b^2$$$$(2):\,\left(\frac{\sin\theta}{\cos^2\theta}\right)A-\left(\frac{\cos\theta}{\sin^2\theta}\right)B=0$$
you can use the usual tricks to eliminate one of the variables.
for example, if you do ##(1)+\frac{\sin\theta}{\cos\theta}(2)##, you will get some expression in ##A## only.

sahilmm15
After ##cot\theta = {\frac {ax^\frac{1}{3}} {by^\frac{1}{3}}}##, let us assume a right angled triangle. From ##cot\theta## we get the base as ## ax^\frac{1}{3} ## and perpendicular ## by^\frac{1}{3} ##. From Pythagoras theorem we get the hypotenuse as ## \sqrt{ (ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} }##. Now we have $${\frac{ax}{cos\theta}} + {\frac{by}{sin\theta}} = a^2-b^2$$. We apply the values of ##cos\theta## and ##sin\theta## from the right angled triangle respectively. After simplification we get ## \sqrt{ (ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} }\cdot [(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}}]= a^2 - b^2##. Now we know that ##\sqrt{x}\cdot x=x^\frac{3}{2}.##. So $${ [(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}}]^{\frac{3}{2}} } = a^2-b^2$$. Hence we get $$(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} = (a^2-b^2)^{\frac{2}{3}}$$ Difficult question but worth a try.

anuttarasammyak
Gold Member
You choose the way to delete ##\theta## and I choose the way to delete ##by## to get the same result.

sahilmm15