Question

Separate the real & imaginary part
Sec(x+iy)

Answer 1

Step-by-step explanation:

sec(x+iy)=cosec[90-(x+iy)]

comparing the equation

x+iy=90-x-iy

2(iy)=90-2x

iy=2(45-x)÷2

I=45-x÷y

Answer 2

Real part = $\dfrac{2\cos x\cos hy}{\cos 2x+\cosh 2y}$ and Imaginary part = $\dfrac{2\sin x\sin hy}{\cos 2x+\cosh 2y}$

Step-by-step explanation:

We have,

$\sec(x+iy)$

To separate the real and imaginary part of $\sec(x+iy)$ = ?

∴ $\sec(x+iy)$

$=\dfrac{1}{\cos(x+iy)}$

Rationalising numerator and denominator, we get

$=\dfrac{1}{2\cos(x+iy)}\times \dfrac{2\cos(x-iy)}{\cos(x-iy)}$

$=\dfrac{2(\cos x\cos iy+\sin x\sin iy)}{\cos 2x+\cos 2iy}$

Using the trigonometric identity,

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$=\dfrac{2(\cos x\cos hy+i\sin x\sin hy)}{\cos 2x+\cosh 2y}$

Separating real and imaginary part, we get

$=\dfrac{2\cos x\cos hy}{\cos 2x+\cosh 2y} +i\dfrac{2\sin x\sin hy}{\cos 2x+\cosh 2y}$

∴ Real part = $\dfrac{2\cos x\cos hy}{\cos 2x+\cosh 2y}$ and

Imaginary part = $\dfrac{2\sin x\sin hy}{\cos 2x+\cosh 2y}$