Question

Express 1/1+cos theta -isin theta in the form of a+ib.

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

Answer:

$\frac{1}{1+\cos \theta-i\sin \theta}=\frac{1+\cos \theta}{(1+\cos \theta)^2+(\sin \theta)^2}+i\frac{\sin \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

$a=\frac{1+\cos \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

$b=\frac{\sin \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

Step-by-step explanation:

Given the expression

$\frac{1}{1+\cos \theta-i\sin \theta}$

we have to express the above in the form a+ib

$\frac{1}{1+\cos \theta-i\sin \theta}$

By rationalizing

$\frac{1}{(1+\cos \theta)-i\sin \theta}\times \frac{(1+\cos \theta)+i\sin \theta}{(1+\cos \theta)+i\sin \theta}$

$=\frac{(1+\cos \theta)+i\sin \theta}{(1+\cos \theta)^2-(i\sin \theta)^2}$

$=\frac{(1+\cos \theta)+i\sin \theta}{(1+\cos \theta)^2-i^2(\sin \theta)^2}$

$=\frac{(1+\cos \theta)+i\sin \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

$=\frac{1+\cos \theta}{(1+\cos \theta)^2+(\sin \theta)^2}+i\frac{\sin \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

which is required form where

$a=\frac{1+\cos \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$

$b=\frac{\sin \theta}{(1+\cos \theta)^2+(\sin \theta)^2}$