Question

Prove that:

\[\frac{\cos3A-cosA}{\sin3A-sinA}+\frac{\cos2A-\cos4A}{\sin4A-\sin2A}=\frac{sinA}{\cos2Acos3A}\]

Answer 1

namaste aunty :|

Answer 2

formulae of cosa-cob and sina - sinb should help you.

Answer 3

aunty ji ? there?

Answer 4

thx.........................................................................................................................................................................................................................................................................................................................^infinity...... count d no of dots......... just joking

Answer 5

glad to help.
(not joking)

Answer 6

LOL

Answer 7

\[\frac{\cos3A-cosA}{\sin3A-sinA}+\frac{\cos2A-\cos4A}{\sin4A-\sin2A}=\frac{sinA}{\cos2Acos3A}\] Mark numerator of the first term is (1) , denominator of the first term is (2) , numerator of the second term is (3) , denominator of the second term is (4) . Now, I calculate them, one by one, and then combine later.

Answer 8

\[(1)= cos 3A -cos A = 2sin\frac{3A+A}{2}sin{\frac{A-3A}{2}}\\=2sin 2A sin (-A)\\=-2sin 2Asin A\] because sin (-A)= -sin A

Answer 9

\[(2)=sin 3A - sin A\\=2sin{\frac{3A-A}{2}}cos\frac{3A+A}{2}\\=2sin Acos2A\]

Answer 10

\[\frac{(1)}{(2)}=\frac{-2sin2AsinA}{2sinAcos2A}\\=\frac{-sin2A}{cos2A}\]mark it as (5)

Answer 11

\[(3)=cos2A-cos4A=2sin{\frac{2A+4A}{2}}sin\frac{4A-2A}{2}\\=2sin3AsinA\]

Answer 12

\[(4)=sin4A-sin2A= 2sin{\frac{4A-2A}{2}}cos\frac{4A+2A}{2}\\=2sinAcos3A\]

Answer 13

\[\frac{(3)}{(4)}=\frac{2sin3AsinA}{2sinAcos3A}\\=\frac{sin3A}{cos3A}\]mark it as (6)

Answer 14

(5)+(6) I rearrange the order:
\[\frac{sin3A}{cos3A}-\frac{sin2A}{cos2A}\]
\[=\frac{sin3Acos2A- sin2Acos3A}{cos2Acos3A}\]
the numerator is the form of sin (a -b) = sina cosb -sinb cos a with a = 3A and b = 2A
therefore, it is = sin (3A-2A)= sin A
BINGO, now combine you have
\[\frac{sinA}{cos2Acos3A}\] it's the RHS, right? hehehee.... fiiiiiiiiinaaaaaaallllyyyy