Question

Prove that sin(u+iv)=x+iy then prove x^2/cos^2x -y^2/cos^2u=1

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{sin(u+i\,v)=x+i\,y}$

$\underline{\textbf{To prove:}}$

$\mathsf{\dfrac{x^2}{sin^2u}-\dfrac{y^2}{cos^2v}=1}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{sin(u+i\,v)=x+i\,y}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{Sin(A+B)=sinA\,cosB+cosA\,sinB}}$

$\mathsf{sin\,u\;cos\,iv+cos\,u\;sin\,iv=x+i\,y}$

$\mathsf{Using,}$

$\boxed{\mathsf{sin\,ix=i\;sinhx\;\;\&\;\;cos\,ix=coshx}}$

$\mathsf{sin\,u\;cosh\,v+cos\,u\;i\,sinh\,v=x+i\,y}$

$\mathsf{sin\,u\;cosh\,v+i\;cos\,u\;sinh\,v=x+i\,y}$

$\textsf{Separating real and imaginary parts, we get}$

$\mathsf{sinu\;cosh\,v=x\;\;\;\&\;\;\;cosu\;sinh\,v=y}$

$\mathsf{cosh\,v=\dfrac{x}{sinu}\;\;\;\&\;\;\;sinh\,v=\dfrac{y}{cosu}}$

$\textsf{We know that,}$

$\mathsf{cosh^2v-sinh^2v=1}$

$\mathsf{\left(\dfrac{x}{sinu}\right)^2-\left(\dfrac{y}{cosu}\right)^2=1}$

$\implies\boxed{\mathsf{\dfrac{x^2}{sin^2u}-\dfrac{y^2}{cos^2u}=1}}$

$\underline{\textbf{Find more:}}$

X/acosθ+y/bsinθ=1 and x/asinθ-y/bcosθ=1, prove that x²/a²+y²/b²=2.

https://brainly.in/question/15922910