---
title: "Euler's theorem for cosec⁻¹ function"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:euler-s-theorem-cosec-inverse"
source: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-cosec-inverse
site: BCT Engineering
---

# Euler's theorem for cosec⁻¹ function

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2071 Magh, 5 marks
- Video: [Euler's Theorem Example | Euler's Theorem Question | Euler's Theorem Problems | Deduction of Eule...](https://www.youtube.com/watch?v=mUPbYS11HrY) — FEARLESS INNOCENT MATH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-cosec-inverse
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

If u = cosec⁻¹((x^(1/2)+y^(1/2))/(x^(1/3)+y^(1/3))), show that x ∂u/∂x + y ∂u/∂y = −(1/6) tan u.

## Short answer

Let z = cosec u; it is homogeneous of degree 1/6. By Euler’s theorem x z_x + y z_y = (1/6)z. Substituting z=cosec u yields −cosec u cot u (x u_x + y u_y) = (1/6) cosec u, so x u_x + y u_y = −(1/6) tan u.

## Answer

#### Euler's theorem
If f(x, y) is homogeneous of degree *n*, then x ∂f/∂x + y ∂f/∂y = n f.

#### Proof
Given u = cosec⁻¹((x^(1/2) + y^(1/2))/(x^(1/3) + y^(1/3))). Let

z = cosec u = (x^(1/2) + y^(1/2))/(x^(1/3) + y^(1/3)).

The numerator has degree 1/2 and the denominator has degree 1/3: replacing x, y by tx, ty gives z(tx, ty) = t^(1/2 − 1/3) z = t^(1/6) z. So z is homogeneous of degree n = 1/6.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = (1/6) z

With z = cosec u: ∂z/∂x = −cosec u cot u ∂u/∂x (and similarly for y). Hence

```text
−cosec u cot u ( x ∂u/∂x + y ∂u/∂y ) = (1/6) cosec u
x ∂u/∂x + y ∂u/∂y = −(1/6) / cot u
                   = −(1/6) tan u
```

Hence proved. ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Euler's theorem is applied to which function?
   A. z = u
   B. z = cosec u
   C. z = cot u
   D. z = sin u
   Answer: B. z = cosec u = (x^(1/2)+y^(1/2))/(x^(1/3)+y^(1/3)) is homogeneous. (Key point: Function)

2. What is its degree?
   A. 1/2
   B. 1/6
   C. 1/3
   D. 5/6
   Answer: B. t^(1/2)/t^(1/3) = t^(1/6). (Key point: Degree)

3. Using z_x = −cosec u cot u u_x, the result is:
   A. −(1/6) tan u
   B. −(1/6) cot u
   C. −6 tan u
   D. (1/6) cot u
   Answer: A. -cosec u cot u (x u_x + y u_y) = (1/6) cosec u, so the sum = -(1/6)/cot u = -(1/6) tan u. (Key point: Final result)

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