find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 ncert cbse chapter 8 binomial theorem miscellaneous exercise    

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

using expansion with 

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

(x-y)^4 = (x^4)-4(x^3)(y)+6(x^2)(y^2)-4(x)(y^3)+(y^4)

adding the two expansions 

(x+y)^4 +(x-y)^4= 2[(x^4) +6(x^2)(y^2) +(y^4) ]

use x =(a^2) y = sqrt{(a^2)-1}

[(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 = 2[{(a^2)^4}+6{(a^2)^2}{[ sqrt{(a^2)-1}]^2} +{[ sqrt{(a^2)-1}]^4}]

 =2[ (a^8) +6(a^4)((a^2)-1) +{(a^2)-1}^2 ]

=2 [ (a^8) +6(a^6) -6(a^4) +{(a^4) -2(a^2)+1} ]

=2[ (a^8) +6(a^6) -5(a^4)-2(a^2)+1  ]

=2(a^8) +12(a^6) -10(a^4)- 4(a^2)+2

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

exercise 8.2

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

solution  

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tan15 degrees and sin 75 degrees

tan15 degrees and sin 75 degrees

11th cbse ncert trigonometry

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

using trigonometry formula trigonometry identities

sin(A+B) =sinAcosB+cosAsinB


choose A=45degrees B=30degrees

sin(75degrees)
sin(45+30) =sin45cos30+cos45sin30 ( angles in degree )

=[1/(sqrt(2)][sqrt(3)/2] +[1/(sqrt(2)][1/2]

= {[sqrt(3) +1} / {2*sqrt(2)}

using trigonometry formula trigonometry identities 

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

choose A=45degrees B=30degrees

tan 15degrees

tan[45-30] = [tan45 -tan30]  / [1+tan45tan30] ( angles in degree )

=[1-{1/sqrt(3)}] / [1+(1){1/sqrt(3)}]

=[ sqrt(3) -1 ] / [sqrt(3)} + 1] introduce conjugate

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3) + 1] [ sqrt(3) -1 ] }

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3)}^2 - 1] }
 using identities (a-b)^2 and (a+b)(a-b)
={ [sqrt(3)]^2 - 2*sqrt(3) + 1 } / {3 - 1}

={3-2*sqrt(3) +1} / [2 ]

= [ 4-2*sqrt(3) ] / 2

={ 2 * [2- sqrt(3)]} / 2

= [2 - sqrt(3)]

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

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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exercise 3.3 ncert trigonometry 12

exercise 3.3 ncert trigonometry 12

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

using trigonometry formula trigonometry identities

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

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

finally use 2sinxcosx  = sin2x

Factorise and then apply the identities

LHS = {sin6x + sin4x } {sin6x - sin4x}

={2sin(10x/2)cos(2x/2)}{2cos(10x/2)sin(2x/2)}

={2sin5xcosx}{2cos5xsinx}

={ 2sinxcosx}{2sin5xcos5x} on regrouping

= {sin2x} {sin10x} using sin2x formula

=RHS

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

using trigonometry formula trigonometry identities

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

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

finally use 2sinxcosx  = sin2x

LHS =(cos2x)^2  - (cos6x)^2

={cos2x+cos6x} {cos2x-cos6x}

= { -2sin(8x/2)sin(-4x/2)} {2cos(8x/2)cos(-4x/2)}

= {-2sin4xsin(-2x)}{2cos4xcos(-2x)} {using sin(-x)= -sinx and cos(-x)=cosx}

={-2sin4x[-sin2x]}{2cos4xcos2x}

={2sin2xcos2x}{2sin4xcos4x} on regrouping

=sin4xsin8x

=RHS.

 

3.3

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

solution

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

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
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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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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exercise 3.3 ncert trigonometry

exercise 3.3 ncert trigonometry cbse 11th

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

using trigonometry formula trigonometry identities

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

(cosx)^2 - (sinx)^2 = cos2x in denominator


LHS = [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] introducing (-1) to change the
order in both numerator and denominator

=[sin3x - sinx] / [ (cosx)^2 - (sinx)^2 ]

= [2cos(4x/2) sin(2x/2)]  / [cos2x]

= [2cos(2x) sin(x)]  / [cos2x]

= 2 sinx = RHS





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

using trigonometry formula trigonometry identities

tan2x = { 2 tanx }  / { 1 - [(tanx)^2]}
Replace x with 2x

LHS = tan4x =  tan{2(2x)}

= { 2 tan2x }  / { 1 - [(tan2x)^2]} again using the same formula

={ 2[{ 2 tanx }  / { 1 - [(tanx)^2]}] }/ { 1 -[{ 2 tanx }  / { 1 - [(tanx)^2]}]^2 }

expand using identity for (a-b)^2 in denominator . . .

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




3.3

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


miscellaneous

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work