Friday, July 21, 2017

Q. Prove that: (cos5a.sin2a-cos4a.sin3a)/(sin5a.sin2a-cos4a.cos3a) = cot2a

Soln:
L.H.S.
= (cos5a.sin2a-cos4a.sin3a)/(sin5a.sin2a-cos4a.cos3a)
Multiply numerator and denominator by 2.
= 2(cos5a.sin2a - cos4a.sin3a) / 2(sin5a.sin2a - cos4a.cos3a)
=  (2cos5a.sin2a - 2cos4a.sin3a) / (2sin5a.sin2a - 2cos4a.cos3a)
= [sin(5a+2a)-sin(5a-2a)-sin(4a+3a)+sin(4a-3a)]/[cos(5a-2a)-cos(5a+2a)-sin(4a-3a)+cos(4a+3a)]
= (sina - sin3a)/(cso3a-cosa)
= (- 2cos2a.sina)/(-2sin2a.sina)
= cos2a/sin2a
= cot2a
= R.H.S.


Formulae:

sin(A+B) - sin(A-B) = 2cosA.sinB
sinC - sinD = 2cos[(C+D)/2].cos[(C-D)/2]
cosC - cosD = -2sin[(C+D)/2].cos[(C-D)/2]