Math, asked by nayan7228, 1 year ago

sin x- sin3x+sin5x - sin7x/
cos x-cos3x-cos5x+cos7x
=cot2x​

Answers

Answered by knjroopa
2

Step-by-step explanation:

Given sin x- sin3x+sin5x - sin7x/ cos x-cos3x-cos5x+cos7x = cot2x

  • Given sin x- sin3x+sin5x - sin7x/ cos x-cos3x-cos5x+cos7x =cot2x
  •              We can write this as    Sinx – sin3x + sin5x – sin7x / cosx – cos3x – cos5x + cos7x
  •                  (sinx – sin 3x) + (sin 5x – sin 7x) / (cos x – cos 3x) – (cos 5x – cos 7x)
  •          2 cos (x + 3x) sin (x – 3x) / 2 + 2 cos (5x + 7x) / 2 sin (5x – 7x) / 2 /        2 sin (x + 3x)/2 sin (3x – x) / 2 – 2 sin (5x + 7x)/ 2 sin(7x – 5x) / 2
  •                     2 cos2x (- sin x) + 2cos6x(- sin x) / 2sin 2x sin x – 2 sin 6x sin x
  •                         = - 2sinxcos 2x – 2 sinx cos 6x / 2sin 2x sin x – 2 sin 6x sin x
  •                       = - 2sinx (cos 2x + cos 6x) / 2 sinx (sin 2x – sin 6x)
  •                         After cancelling 2 sinx  
  •                      = we get  - (cos 2x + cos 6x) / (sin 2x – sin 6x)
  •                      = - (2 cos 2x + 6x / 2 cos 2x – 6x / 2) / 2 cos 2x + 6x / 2 sin 2x – 6x / 2
  •                           cos (- θ) = cos θ
  •                       = - cos 2x / - sin 2x
  •                        = cot 2x

Reference link will be

https://brainly.in/question/12416155

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