Question

sina-sin5a+sin9a-sin13a/cosa-cos5a-cos9a+cos13a=cot4a. prove​

Answer 1

Answer:

Step-by-step explanation:

Sina-sin5a+sin9a-sin13a/cosa-cos5a-cos9a+cos13a

=> (sina - sin13a) + (sin9a - sin5a)/(cosa + cos13a) -  (cos9a + cos5a)

But sinA - sinB = 2cos(A+B/2)sin(A-B/2)

     cosA + cosB = 2cos(A+B/2)cos(A-B/2)

=> [2cos(a+13a/2)sin(a-13a/2) + 2cos(9a+5a/2).sin(9a-5a/2)] / 2cos(a+13a/2)cos(a-13a/2) - 2cos(9a+5a/2)cos(9a-5a/2) +

=> 2cos7a sin(-6a) + 2cos7asin2a / 2cos7acos(-6a) - 2cos7acos2a

= 2cos7a [-sin6a+sin2a]/2cos7A[cos6a - cos2a]

= sin2a - sin6a/cos6a - cos2a

    But sinA - sinB = 2cos(A+B/2)sin(A-B/2)

     cosA -  cosB =  - 2sin(A+B/2)sin(A-B/2)

=> 2cos(2a+6a/2)sin(2a-6a/2) / - 2sin(6a+2a/2)sin(6a-2a/2)

=> 2cos4asin(-2a) / - 2sin4asin2a

=> - 2cos4asin2a/ - 2sin4asin2a

=> cos4a/sin4a

=> Cot4a.

=> R.H.S

Hence proved.