show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
Question
Maths
July 15, 2018, 12:23 p.m.
Answer
show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
Maths
July 15, 2018, 12:23 p.m.
Answer
show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
Maths
July 15, 2018, 12:23 p.m.
Answer
show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
Maths
July 15, 2018, 12:23 p.m.
Answer
show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
Maths
July 15, 2018, 12:23 p.m.
Answer
show that
sin [n+1] x cos [n+2]x +cos [ n+1] x cos [n+1]x.cos [n+2] =sin [x+y].
.
if a belongs to set of -1,2,3,4,5, and b belongs to set of 0,3,6 form the set of all ordered pairs (a,b) such that a+b=5
Accounts Tax GST
July 15, 2018, 8:08 a.m.
Answer
i am working as a contract employee.my employer had deducted already 10%tds-my gross annual income 740000 .any possibility of gaining refund
derection determine whether the relation represented by the equaltion is a fuction
Sir can you please tell me that in chapter surface areas and volumes exercise no.13.2 . Question no.4 how did you take area of 1 revolution=area of roller....i have not understood only this single line...can you please tell my answer
