LOADING . . .

CBSE Solutions for Class 11 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Prove that:

cos 80° + cos 40° − cos 20° = 0

Hide | Show

જવાબ : Consider LHS:cos 80° + cos 40° − cos 20° = 2cos (80°/2 + 40°/2) cos (80°/2 − 40°/2) − cos 20°                   {∵ cosA+cosB=2cos(A/2+B/2)cos(A/2−B/2)} = 2cos 60° cos 20°− cos 20° = 2× ½ cos 20° − cos20° = cos 20° − cos 20°= 0 Hence, LHS=RHS.


Prove that:

sin 50° − sin 70° + sin 10° = 0

Hide | Show

જવાબ : Consider LHS:sin 50° − sin 70° + sin 10° = 2sin (50°/2 −70°/2) cos (50°/2 + 70°/2)  + sin 10°                 {∵ sin A − sin B = 2sin (A/2-B/2) cos (A/2 + B/2)} = 2sin (−10°) cos 60° + sin 10° =2× ½ sin (−10°) + sin 10° = −sin 10°+sin 10°= 0 Hence, LHS=RHS.


Prove that:

cos 55° + cos 65° + cos 175° = 0

Hide | Show

જવાબ : Consider LHS: cos 55° + cos 65° + cos 175° = 2cos (55°/2 + 65°/2) cos (55°/2 − 65°/2) + cos 175°               {∵ cosA+cosB=2cos(A/2+B/2)cos(A/2−B/2)} = 2cos 60° cos(−5°) + cos 175° = 2× ½ cos 5° + cos 175° =cos 5° + cos 175° = 2cos (5°/2 + 175°/2) cos (5°/2 − 175°/2) = 2cos 90° cos 85°= 0 Hence, LHS=RHS.


Prove that:

sin 40° + sin 20° = cos 10°

Hide | Show

જવાબ : Consider LHS:sin 40° + sin 20° = 2sin (40°/2 + 20°/2) cos (40°/2 − 20°/2)                   {∵ sinA+sinB=2sin(A/2+B/2)cos(A/2−B/2)} = 2sin 30° cos 10°= 2× ½ cos 10° = cos(10°) Hence, LHS = RHS


Prove that:

sin 105° + cos 105° = cos 45°

Hide | Show

જવાબ : Consider LHS:sin 105° + cos 105°= sin 105° + cos (90° + 15°)= sin 105° − sin 15°= 2sin (105°/2 − 15°/2) cos (105°/2 + 15°/2)    {∵ sinA+sinB=2sin(A/2−B/2)cos(A/2+B/2)} = 2sin 45°cos 60°= 2sin (90° − 45°) cos 60° = 2× ½ cos(45°) =cos 45° Hence, LHS=RHS.


Prove that:

sin 23° + sin 37° = cos 7°

Hide | Show

જવાબ : Consider LHS:sin 23° + sin 37° = 2sin (23°/2 + 37°/2) cos (23°/2− 37°/2)                       {∵ sin A + sin B = 2sin (A/2 + B/2) cos (A/2 – B/2)} = 2sin 30° cos (−7°) = 2sin 30°cos 7°= 2× ½ cos 7°= cos 7° Hence, LHS=RHS.


Prove that:

sin 50° + sin 10° = cos 20°

Hide | Show

જવાબ : Consider LHS:sin 50° + sin 10° = 2sin (50°/2 + 10°/2) cos (50°/2 − 10°/2)                         {∵ sin A + sin B = 2sin (A/2 + B2/) cos (A/2 – B/2)} = 2sin 30° cos 20°=2× ½ cos 20°= cos 20° Hence, LHS = RHS.


Prove that:

cos 100° + cos 20° = cos 40°

Hide | Show

જવાબ : Consider LHS:cos 100° + cos 20°= 2cos (100°/2 + 20°/2) cos (100°/2 − 20°/2)                            {∵ cosA+cosB=2cos(A/2+B/2)cos(A/2−B/2)} = 2cos 60° cos 40° = 2× ½ cos 40° = cos 40° Hence, LHS=RHS.


Express each of the following as the product of sines and cosines : sin 2x + cos 4x

Hide | Show

જવાબ : 2sin (π/4 − x) cos (3x – π/4)


Find the radian measure corresponding to the following degree measures : −47° 30'

Hide | Show

જવાબ : -19π/72 rad


Find the radian measure corresponding to the following degree measures : 300°
 

Hide | Show

જવાબ : 5π/3 rad


Find the radian measure corresponding to the following degree measures : 7° 30'

Hide | Show

જવાબ : π/24 rad


Find the radian measure corresponding to the following degree measures : 125° 30'

Hide | Show

જવાબ : 251π/360 rad


Find the radian measure corresponding to the following degree measures : −300°

Hide | Show

જવાબ : -5π/3 rad


Find the radian measure corresponding to the following degree measures : 35°

Hide | Show

જવાબ : 7π/36 rad


Find the radian measure corresponding to the following degree measures : −56°

Hide | Show

જવાબ : - 14π/45 rad


Find the radian measure corresponding to the following degree measures : 135°

Hide | Show

જવાબ : 3π/4 rad


Find the values of the following trigonometric ratios : sin(5π/3)

Hide | Show

જવાબ : -√3 / 2


Find the values of the following trigonometric ratios : sin 17π

Hide | Show

જવાબ : 0


Find the values of the following trigonometric ratios : tan(11π/6)
 

Hide | Show

જવાબ : -1/√3


Find the values of the following trigonometric ratios : sin(151π/6)

Hide | Show

જવાબ : -1/2


Find the values of the following trigonometric ratios : cos(39π/4)

Hide | Show

જવાબ : 1/√2


Find the values of the following trigonometric ratios : sin(41π/4)

Hide | Show

જવાબ : 1/√2


Find the values of the following trigonometric ratios : cos(-25π/4)

Hide | Show

જવાબ : 1/√2


Find the values of the following trigonometric ratios :  tan(7π/4)

Hide | Show

જવાબ : -1


Find the values of the following trigonometric ratios : sin(17π/6)

Hide | Show

જવાબ : 1/2


Find the values of the following trigonometric ratios : cos(19π/6)

Hide | Show

જવાબ : -√3/2


Find the values of the following trigonometric ratios : sin(-11π/6)

Hide | Show

જવાબ : 1/2


Find the values of the following trigonometric ratios :  cos(19π/4)

Hide | Show

જવાબ : -1/√2


Find the values of the following trigonometric ratios :  tan(-13π/4)

Hide | Show

જવાબ : -1


Find the values of the following trigonometric ratios : cosec(-20π/3)
 

Hide | Show

જવાબ : -2/√3


Prove that :  cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Hide | Show

જવાબ : LHS =cos (570°)sin (510°) + sin (−330°)cos (−390°) =cos (570°) sin (510°) + [−sin (330°)]cos (390°)                        [∵ sin(−x) = −sin x and cos(−x) = cos x] =cos (570°)sin(510°) −sin (330°) cos (390°) =cos (90°×6+30°) sin (90°×5+60°) −sin (90°×3+60°) cos (90°×4+30°) =−cos (30°) cos (60°) −[−cos (60°)] cos (30°) =−cos (30°) cos (60°) +cos (30°) sin (60°) = 0 = RHS


Prove that :  tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0

Hide | Show

જવાબ : LHS = tan (−225°) cot (−405°) − tan (−765°) cot (675°)                 =[− tan (225°)][−cot (405°)] − [−tan (765°)] cot (675°)     [∵ tan (−x) = tan (x) and cot (−x) = −cot (x)] = tan (225°) cot (405°) +tan (765°) cot (675°) =tan (90°×2+45°) cot (90°×4+45°) + tan (90°×8+45°) cot (90°×7+45°) =tan (45°) cot (45°) + tan (45°)[−tan (45°)] = 1×1 + 1×(−1) = 1−1 = 0 = RHS


Prove that :  cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = ½

Hide | Show

જવાબ : LHS = cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = cos 24° + cos (90°−35°) + cos (90°×1+35°)+ cos (90°×2+24°) + cos (90°×3+30°) =cos 24° + sin 35° − sin 35°− cos 24° + sin 30° =0 + 0+ 12 = 12 = RHS


Prove that :  tan 225° cot 405° + tan 765° cot 675° = 0

Hide | Show

જવાબ : LHS = tan225°cot405° + tan765°cot675° =tan (90°×2+45°)cot (90°×4+45°) + tan (90°×8+45°) cot (90°×7+45°) =tan (45°) cot (45°) + tan (45°)[−tan (45°)] = 1×1 + 1×(−1) = 1−1  = 0 = RHS


Find the general solutions of the following equations : sin 3x+cos 2x=0

Hide | Show

જવાબ : sin3x + cos2x = 0     ⇒cos2x =- sin3x⇒ cos2x= cos(π/2 + 3x)⇒ 2x= 2nπ ± (π/2 + 3x), n ∈Z
    On taking positive sign, we have:
     2x = 2 + π/2 + 3x⇒ -x = 2 + π/2⇒ x = 2 – π/2, m=-n ∈ Z⇒ x = (4m -1)π/2, m∈ Z
   On taking negative sign, we have:
    2x = 2 – π/2 - 3x⇒ 5x = 2 – π/2⇒ x = (4n -1)π/10, nZ


Find the general solutions of the following equations : sin x=tan x

Hide | Show

જવાબ : sinx = tanx
     ⇒ sinx - tanx = 0⇒ sinx – sinx/cosx = 0⇒ sinx (1 – 1/cosx) = 0⇒sinx (cosx -1) = 0
    
     ⇒ sinx = 0  or  cosx - 1 = 0
   Now,  
sinx= 0 ⇒x = nZ

cosx - 1 = 0 ⇒ cosx = 1 ⇒ cosx= cos0 ⇒ x= 2m ∈Z


Find the general solutions of the following equations : sin 2x+cos x=0

Hide | Show

જવાબ : sin2x + cosx= 0
    ⇒cosx=-sin 2x⇒ cosx= cos (π/2 + 2x)⇒x= 2 ± (π/2 + 2x), nZ
On taking positive sign, we have:
    x= 2 + π/2 + 2x⇒-x = 2  + π/2⇒ x= 2 – π/2, m =-n ∈Zx = (4m -1)π/2, mZ
On taking negative sign, we have:
x= 2nπ – π/2 - 2x⇒ 3x = 2 – π/2⇒ x =(4n - 1)π/6, n∈ Z


Find the general solutions of the following equations : tan px=cot qx

Hide | Show

જવાબ : tanpx =cotqx
     ⇒tanpx= tan (π/2 - qx)⇒ px =  + (π/2 - qx  ), n∈ Z⇒ (p + q)x =  + π/2, n∈ Zx= (2n + 1p + q)π/2, nZ


Find the general solutions of the following equations : tanmx + cot nx=0

Hide | Show

જવાબ : tanmx + cotnx = 0
   ⇒ tanmx =-cotnx⇒ tanmx = tan (π/2 + nx)⇒ mx =  + (π/2 + nx), r ∈Z⇒ (m - nx =  + π/2, r ∈Z⇒ x = (2r + 1m - n)π/2, r ∈Z


Find the general solutions of the following equations : tan 2x tan x=1

Hide | Show

જવાબ : tan2x tanx = 1
    ⇒tan2x =1/tan x⇒ tan2x = cot x⇒ tan2x =tan (π/2 - x)⇒ 2x =  + (π/2 -x), nZ⇒ 3x + π/2, nZxnπ/3 + π/6, nZ


Find the general solutions of the following equations : tan 3x=cot x

Hide | Show

જવાબ : tan3x = cotx
    ⇒ tan3x = tan (π/2 - x)⇒ 3x =  + (π/2 - x), nZ⇒ 4x =  + π/2, nZx = nπ/4 + π/8, nZ


Find the general solutions of the following equations: tanx + cot 2x=0

Hide | Show

જવાબ : tanx + cot2x = 0 
  ⇒ tan x =-cot2x⇒ tanx = tan (π/2 + 2x)⇒ x =  + (π/2 + 2x), nZ⇒ -x =  + π2, nZ⇒ x = -nπ – π/2, n∈Z ⇒ x = mπ – π/2, m = -n∈Z


Find the general solutions of the following equations : sin 2x=cos 3x

Hide | Show

જવાબ : sin2x= cos3x
     cos3x =sin2x
 ⇒ cos3x = cos (π/2-2x)
 ⇒ 3x = 2nπ ± (π/2-2x), nZ
On taking positive sign, we have:
 3x = 2nπ + (π/2-2x)
 ⇒ 5x= 2nπ + π/2
⇒ x = 2nπ/5 + π/10
 ⇒ x= (4n + 1)π/10, n ∈ Z
 Now, on taking negative sign, we have:
 3x = 2nπ – π/2 + 2xnZ
 ⇒  x= 2nπ – π/2
 ⇒ x = (4n - 1)π/2, n∈ Z 


Find the general solutions of the following equations: sin 9x=sin x

Hide | Show

જવાબ : sin9x = sinx
       sin9x - sinx= 0
        2 sin (9x/2 – x/2) cos (9x/2 + x/2)= 0
       sin 8x/2 = 0 or cos 10x2 = 0
       sin 4x= 0 or cos 5x= 0
       4xnπ, n Z or 5x = (2n + 1)π/2, n Z
      xnπ/4,  n Z or x= (2n + 1)π/10, n Z


Express each of the following as the product of sines and cosines : sin 12x + sin 4x

Hide | Show

જવાબ : 2sin8xcos4x


Express each of the following as the product of sines and cosines : sin 5x − sin x

Hide | Show

જવાબ : 2sin2x cos3x


Express each of the following as the product of sines and cosines : cos 12x + cos 8x

Hide | Show

જવાબ : 2 cos 10x cos 2x


Express each of the following as the product of sines and cosines : cos 12x − cos 4x

Hide | Show

જવાબ : -2sin A + B2 sin A - B2=-2 sin 8x sin 4x


Express each of the following as the product of sines and cosines : sin 2x + cos 4x

Hide | Show

જવાબ : 2sin (π/4 − x) cos (3x – π/4)


Write the value of 2 (sin6 x + cos6 x) −3 (sin4 x + cos4 x) + 1.

Hide | Show

જવાબ : 2(sin6x+cos6x)-3(sin4x+cos4x)+1= 2(sin2x+cos2x)(sin4x+cos4x-sin+x.cos+x)-3(sin4x+cos4x)+1

=2.1(sin4x+cos4x-sin2x.cos2x)-3(sin4x+cos4x)+1

=2(sin4x+cos4x)-2sin2x.cos2x-3(sin4x+cos4x)+1=-(sin4x+cos4x)-2sin2x.cos2x+1

=-{sin4x+cos4x+2sin2x.cos2x}+1

=-(sin2x+cos2x)2+1=-1+1=0


Write the value of cos 1° + cos 2° + cos 3° + ... + cos 180°.

Hide | Show

જવાબ : cos1°+cos2°+cos3°+...+cos180°

=cos1°+cos2°+cos3°+...+cos88°+cos89°+cos90°+cos(180-89)°+cos(180-88)°+...+cos(180-1)°+ cos180°      [cos(180°-θ)=-cos θ]

=cos1°+cos2°+cos3°+...+cos88°+cos89°+cos90°-cos89°-cos88°-...-cos1°+ cos180°=cos90°+cos180°=0-1=-1


If cot (α + β) = 0, then write the value of sin (α + 2β).

Hide | Show

જવાબ : cot (α+β)=0α+β=π/2              (1)

β=π/2-α                   (2)

α=π/2-β                   (3)

Now, sin(α+2β) = sin(α+β+β)

 =sin(π/2+π/2-α)

=sin(π-α)

=sin α

Now, sin(α+2β) = sin(α+2β)                             

=sin(π/2-β+2β)

=sin(π/2+β)

=cos β


If tan A + cot A = 4, then write the value of tan4 A + cot4 A.

Hide | Show

જવાબ : tanA+cotA=4

Squaring both the sides:tan2A+cot2A+2=16

tan2A+cot2A=14

Squaring both the sides again:tan4A+cot+A+2=196

tan4A+cot4A=194


Write the least value of cos2 x + sec2 x.

Hide | Show

જવાબ : We know:
cos x can take the minimum value of -1.

cos2 x + sec2 x

=cos4x+1/cos2x=(-1)4+1(-1)2=2


Sketch the graphs of the following functions:
f(x) = 2 cosec πx

Hide | Show

જવાબ : f(x) = 2 cosec πx


Sketch the graphs of the following functions:
f(x) = 3 sec x

Hide | Show

જવાબ : f(x) = 3 sec x


Sketch the graphs of the following functions:
f(x) = cot 2x

 

Hide | Show

જવાબ : f(x) = cot 2x


Sketch the graphs of the following functions:
f(x) = tan2 x

 

Hide | Show

જવાબ : f(x) = tan2 x


Sketch the graphs of the following functions:
f(x) = 2 sec πx

Hide | Show

જવાબ : f(x) = 2 sec πx


If sin A=4/5and cos B=51/3, where 0 < A, B<π2, find the values of the following:

cos (A − B)

Hide | Show

જવાબ : Given:       sinA = 4/5 and cosB = 51/3

We know that        cosA = √(1 - sin2A)       and   sinB = √(1 - cos2B)          ,  where 0 < A , B < π/2

  cosA = √(1 - (4/5)2)      and    sinB = √[1 - (5/13)2]

  cosA =√[1 – 16/25]   and    sinB = √[1 – 25/169]

  cosA =√9/25    and    sinB = √144/169

  cosA =3/5   and     sinB = 12/13

Now,

cos(A- B) = cosA cosB + sinA sinB                         

=3/5×5/13 + 4/5×12/13                         

=15/65 + 48/65                         

=63/65


If sin A=4/5and cos B=51/3, where 0 < A, B<π2, find the values of the following:

 sin (A − B)

Hide | Show

જવાબ : Given:       sinA = 4/5 and cosB = 51/3

We know that        cosA = √(1 - sin2A)       and   sinB = √(1 - cos2B)          ,  where 0 < A , B < π/2

  cosA = √(1 - (4/5)2)      and    sinB = √[1 - (5/13)2]

  cosA =√[1 – 16/25]        and    sinB = √[1 – 25/169]

  cosA =√9/25                and    sinB = √144/169

  cosA =3/5                      and     sinB = 12/13

Now,sin(A-B)=sinA cosB-cosA sinB                        

=4/5×5/13-3/5×12/13                        

=20/65-36/65                        

=-16/65


If sin A=4/5and cos B=51/3, where 0 < A, B<π2, find the values of the following:

cos (A + B)

Hide | Show

જવાબ : Given:       sinA = 4/5 and cosB = 51/3

We know that        cosA = √(1 - sin2A)       and   sinB = √(1 - cos2B)          ,  where 0 < A , B < π/2

  cosA = √(1 - (4/5)2)      and    sinB = √[1 - (5/13)2]

  cosA =√[1 – 16/25]        and    sinB = √[1 – 25/169]

  cosA =√9/25                and    sinB = √144/169

  cosA =3/5                      and     sinB = 12/13

Now,Given:       sinA = 4/5 and cosB = 5/13

We know that        cosA = √(1 - sin2A)       and   sinB = √(1 - cos2B)          ,  where 0 < A , B < π/2

  cosA = √(1 - (4/5)2)      and    sinB = √[1 - (5/13)2]

  cosA =√[1 – 16/25]        and    sinB = √[1 – 25/169]

  cosA =√9/25                and    sinB = √144/169

  cosA =3/5                      and     sinB = 12/13

Now,cos(A+B) = cosA cosB - sinA sinB                        

=3/5×5/13 – 4/5×12/13                        

=15/65 – 48/55                        

=-33/65


Prove that:
cos 4x=1-8 cos2x+8 cos4 x

Hide | Show

જવાબ : LHS=cos4x      =cos(2×2x)      =2cos2×2x-1   [cos2θ=2cos2θ-1]      

=2(2cos2x-1)2-1 [cos22θ=(2cos2θ-1)2]

      =2(4cos4x-4cos2x+1)-1      =8cos4x-8cos2x+1      =1-8cos2x+8cos4x=RHS

Hence proved.


Prove that:
sin 4x=4 sin x cos3x-4 cos x sin3 x

Hide | Show

જવાબ : LHS=sin 4x      =2sin2x cos2x   (sin2θ=2sinθcosθ)

Now, using the identities sin2α=2sinαcosα and cos2α=cos2α-sin2α, we get

LHS=2(2sinx cosx).(cos2x-sin2x)      =4sinx cos3x-4sin3x cosx=RHS

Hence proved


In triangle ABC, prove the following:
b sin B-c sin C=a sin (B-C)

Hide | Show

જવાબ : Let asinA=b sinB=csinC=k
Then,

Consider the LHS of he equation b sin B-c sin C=a sin (B-C).
LHS=ksinBsinB-ksinCsinC 
       =k(sin2B-sin2C)=k[sin(B+C)sin(B-C)]             [ sin2B-sin2C=sin(B+C)sin(B-C)]

=k[sin(π-A)sin(B-C)]             [ A+B+C=π]

=ksinAsin(B-C)                         [a=ksinA]

=asin(B-C)=RHS

Hence proved.


In triangle ABC, prove the following:
b sin B-c sin C=a sin (B-C)

Hide | Show

જવાબ : Let a/sinA =b/ sinB =c/sinC =k
Then,

Consider the LHS of he equation b sin B-c sin C=a sin (B-C).
LHS=ksinBsinB-ksinCsinC 
       =k(sin2B-sin2C)=k[sin(B+C)sin(B-C)]             [ sin2B-sin2C=sin(B+C)sin(B-C)]

=k[sin(π-A)sin(B-C)]             [ A+B+C=π]

=ksinAsin(B-C)                         [a=ksinA]

=asin(B-C)=RHS

Hence proved.


Sketch the graphs of the following functions:
f(x) = cot2 x

 

Hide | Show

જવાબ : f(x) = cot2 x


Sketch the graphs of the following functions:
f(x)=cotπx2fx=cotπx2

 

Hide | Show

જવાબ : f(x)=cotπx2fx=cotπx2


There are No Content Availble For this Chapter

1

Tan 180

A

  • ½

2

cos 40° + cos 80° + cos 160° + cos 240°

B

½

3

sin 163° cos 347° + sin 73° sin 167°

C

                           0

Hide | Show

જવાબ :

1-C, 2-A, 3-B

1

cos 135

A

1

2

Cos 180

B

-2/2

3

 Cos 360

C

-1

Hide | Show

જવાબ :

1-B, 2-C, 3-A

1

2 π

A

1440°

2

3 π

B

360°

3

4 π

C

720°

4

5 π

D

1080°

Hide | Show

જવાબ :

1-B, 2-C, 3-D, 4-A

Radian to Degree

1

Π

A

240°

2

3 π/2

B

120°

3

2π/3

C

270°

4

4 π/3

D

180°

Hide | Show

જવાબ :

1-D, 2-C, 3-B, 4-A

Radian to Degree

1

π/6

A

60°

2

π/4

B

90°

3

π/3

C

45°

4

π/2

D

30°

Hide | Show

જવાબ :

1-D, 2-C, 3-A, 4-B

Radian to Degree

1

0

A

22.5°

2

Π/12 

B

3

π/8

C

15°

Hide | Show

જવાબ :

1-B, 2-C, 3-A

Degree to Radian

1

180°

A

Π

2

270°

B

3π/2

3

360°

C

Hide | Show

જવાબ :

1-A, 2-C, 3-B

Degree to Radian

1

120°

A

11 π/12

2

135°

B

2 π/3

3

150°

C

9 π/12

4

165°

D

5 π/6

Hide | Show

જવાબ :

1-B, 2-C, 3-D, 4-A

Degree to Radian

1

60°

A

π/2

2

75°

B

π/3

3

90°

C

7 π/12

4

105°

D

5 π/12

Hide | Show

જવાબ :

1-B, 2-D, 3-A, 4-C

Degree to Radian

1

A

π/12

2

15°

B

π/6

3

30°

C

0

4

45°

D

π/4

Hide | Show

જવાબ :

1-C, 2-A, 3-B, 4-D

Download PDF

Take a Test

Choose your Test :

Trignometric Functions

-.

આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.

Browse & Download CBSE Books For Class 11 All Subjects

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.