[ tan{(pi/4)+x} ] / [ tan{(pi/4 )- x} ] = { [1+tanx] / [1-tanx] }^2

ncert  cbse 11th trigonometry exercise 3.3

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2

using trigonometry formula trigonometry identities

cos(pi+x) = -cosx [third quadrant]
cos(-x) = cosx  [fourth quadrant]

sin(pi-x) = sinx  [second quadrant]
cos((pi/2)+x) = -sinx [second quadrant]

LHS =  [cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ]

={ [-cosx][cosx]  } / {[sinx] [-sinx]}

=[cosx/sinx]^2

= [cotx]^2


7.prove that
[ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2

using trigonometry formula trigonometry identities

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

tan(pi/4) = 1



LHS =[ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]

[tan(A+B)] / [tan(A-B)] form with A= (pi/4)  B =x

={ [tan(pi/4)+tanx] / [1 -tan(pi/4)tanx] } / {[tan(pi/4)-tanx] / [1+tan(pi/4)tanx]}

use tan(pi/4) = 1

={[1+tanx]/[1-tanx]} / {[1-tanx] / [1+tanx]}

= { [1+tanx] / [1-tanx] }^2 =RHS


9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1

using trigonometry formula trigonometry identities


cos[(3pi/2)+x] = sinx [fourth quadrant ]

cos[2pi+x] = cosx [first quadrant]

cot[(3pi/2)-x] = tanx [third quadrant]

cot[2pi+x] =cotx

LHS =cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]}

=sinx cosx {tanx +cotx }

=sinx cosx { [sinx/cosx] +[cosx/sinx] }

=sinx cosx { [(sinx)^2 + (cosx)^2]  / [cosx sinx] }

=sinx cosx { [1]  / [cosx sinx] }

=1 =RHS

3.3

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

7. [ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2
solution

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2
 solution

9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1
 solution

 

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

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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