Complex Numbers And Quadratic Equations question 675
Question: $ A+iB $ form of $ \frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan v)} $ is equal to:
Options:
A) $ \sin u\cos v[\cos (x+y-u-v)+ $ $ i\sin (x+y-u-v)] $
B) $ \sin u\cos v[\cos (x+y+u+v)+ $ $ i\sin (x+y+u+v)] $
C) $ \sin u\cos v[\cos (x+y+u+v)- $ $ i\sin (x+y-u+v)] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan v)} $ $ =\frac{(cosx+isinx)(cosy+isiny)}{(cosu+isinu)(cos\nu +isin\nu )} $ $ = sin u cos \nu [cos ( x +y - u - v ) $ $ +i\sin ( x+y-u-\nu )] $
BETA
