ncert 11th cbse trigonometry exercise 3.3 question16

ncert 11th cbse trigonometry exercise 3.3 question16


16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x

using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS = [cos9x -cos5x] / [sin17x - sin3x ]

={ -2sin(14x/2) sin(4x/2) } / { 2cos(20x/2)sin(14x/2) }

= {-2sin7xsin2x} / {2cos10x sin7x}

=  (-sin2x) /cos10x

=RHS

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]

using trigonometry formula trigonometry identities

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS = [sinx -siny] / [cosx +cosy]

={2cos[(x+y)/2] sin [(x-y)/2]} /{2cos[(x+y)/2] cos[(x-y)/2]}

={sin [(x-y)/2]} / { cos[(x-y)/2]}

=tan[(x-y)/2]

=RHS




3.3

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution


19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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