Question

(cos3A - cosA)/(sin 3A - sinA) + cos2A - cos4A/sin4A - sin2A​

Answer 1

Answer:

Step-by-step explanation:

cos3a-cosa / sin3a-sina + cos2a-cos4a / sin4a-sin2a = Sina . secant2a . secant3a

but CosA - CosB = - 2 Sin(a+b/2)Sin(a-b/2)

SinA - SinB = 2 Cos(a+b/2)Sin(a-b/2)

=> - 2sin(3a+a/2)sin(3a-a/2) / 2cos(3a+a/2)sin(3a-a/2)

- 2 sin(2a+4a/2)sin(2a-4a/2) / 2cos(4a+2a/2)sin(4a-2a/2)

= - 2sin2asina/2cos2asina - 2sin3asin(-a)/2cos3asina

= - sin2a/cos2a + sin3a/cos3a

= - sin2a.cos3a + sin3a.cos2a/ cos2a.cos3a

but SinACosB - CosA SinB = Sin(A-B)

= sin (3a -2a) / cos2a.cos3a

= sina/cos2a. cos3a

= sina . secant2a . secant3a

= R.H.S

Hence proved.