11) 2(e2iA)/(e2iA+1)
2) 2(e2iA*e-ipi/2)/(e2iA+1)
3rd one donno
11 Answers
621 +itanA = sec A (cosA +i sin A)
this will be the polar form
R. eiθ
in the first part, R=sec A
Theta=A
now can you try the remaining two parts?
1but bhaiya v can write cos A=(eiA+e-iA)/2..........so sec A itz reciprocal.see#2...arnt v suppose to write dat fully in polar form...or v can leave our answer in term of secA.
12.secA[cos{tan-1(1/tanA)}+i sin{tan-1(1/tanA)}]
3.√{2(1-sinA)}[cos{tan-1(secA+tanA)}+isin{tan-1(secA+tanA)}]
sir,i have noticed the form in which archana gAVE ANS.I NEVER SAW SUCH POLAR FORM IS IT CORRECT.
1nishant sir plz tell the answers to 2nd and 3rd are correct or not,as ans.
is not given in book.
1nishant sir,plz reply whether answer i wrote in#6 is correct as ans. r not given in book.
also,i have noticed the form in which archana gAVE ANS.I NEVER SAW SUCH POLAR FORM IS IT CORRECT
1i have noticed the form in which archana gAVE ANS.I NEVER SAW SUCH POLAR FORM IS IT CORRECT
1plz see #2.
i have noticed the form in which archana gAVE ANS.I NEVER SAW SUCH POLAR FORM IS IT CORRECT
1yes john ma answers r absolutly correct
actually u can write any of the above given form as |z|eia which is called polar form......for example (5cos a+i5 sin a) can b written as 5√2eia got it...& in above questions |z|=1