Prove (x^10)+(y^10) = [(x^5)+(y^5)+(x^2)(y^2)√(2xy)][(x^5)+(y^5)-(x^2)(y^2)√(2xy)]
4:15
Prove (x^10)+(y^10) = [(x^5)+(y^5)+(x^2)(y^2)√(2xy)][(x^5)+(y^5)-(x^2)(y^2)√(2xy)]
14 сағат бұрын
(x-y)(x+y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)][(x^4)-(x^3)y+(x^2)(y^2)-x(y^3)+(y^4)]=(x^10)-(y^10)
3:21
(x-y)(x+y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)][(x^4)-(x^3)y+(x^2)(y^2)-x(y^3)+(y^4)]=(x^10)-(y^10)
14 сағат бұрын
Prove [(x^5)+(y^5)+(x^2)(y^2)√(2xy)][(x^5)+(y^5)-(x^2)(y^2)√(2xy)] = (x^10)+(y^10)
3:40
Prove [(x^5)+(y^5)+(x^2)(y^2)√(2xy)][(x^5)+(y^5)-(x^2)(y^2)√(2xy)] = (x^10)+(y^10)
14 сағат бұрын
Prove (x^9)-(y^9) = (x-y)[(x^2)+xy+(y^2)][(x^6)+(x^3)(y^3)+(y^6)]
4:58
Prove (x^9)-(y^9) = (x-y)[(x^2)+xy+(y^2)][(x^6)+(x^3)(y^3)+(y^6)]
14 сағат бұрын
Prove (x^9)+(y^9) = (x+y)[(x^2)-xy+(y^2)][(x^6)-(x^3)(y^3)+(y^6)]
4:58
Prove (x^9)+(y^9) = (x+y)[(x^2)-xy+(y^2)][(x^6)-(x^3)(y^3)+(y^6)]
14 сағат бұрын
Prove (x-y)[(x^2)+xy+(y^2)][(x^6)+(x^3)(y^3)+(y^6)] = (x^9)-(y^9)
4:30
Prove (x-y)[(x^2)+xy+(y^2)][(x^6)+(x^3)(y^3)+(y^6)] = (x^9)-(y^9)
14 сағат бұрын
Prove (x+y)[(x^2)-xy+(y^2)][(x^6)-(x^3)(y^3)+(y^6)] = (x^9)+(y^9)
4:31
Prove (x+y)[(x^2)-xy+(y^2)][(x^6)-(x^3)(y^3)+(y^6)] = (x^9)+(y^9)
14 сағат бұрын
Prove (x^8)-(y^8) = (x-y)(x+y)[x+y+√(2xy)][x+y-√(2xy)][(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy]
4:21
Prove (x^8)-(y^8) = (x-y)(x+y)[x+y+√(2xy)][x+y-√(2xy)][(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy]
21 күн бұрын
Prove (x^8)+(y^8) = [(x^4)+(y^4)+√2(x^2)(y^2)][(x^4)+(y^4)-√2(x^2)(y^2)]
4:18
Prove (x^8)+(y^8) = [(x^4)+(y^4)+√2(x^2)(y^2)][(x^4)+(y^4)-√2(x^2)(y^2)]
21 күн бұрын
Prove (x-y)(x+y)[x+y+√(2xy)][x+y-√(2xy)][(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy] = (x^8)-(y^8)
4:10
Prove (x-y)(x+y)[x+y+√(2xy)][x+y-√(2xy)][(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy] = (x^8)-(y^8)
21 күн бұрын
Prove [(x^4)+(y^4)+√2(x^2)(y^2)][(x^4)+(y^4)-√2(x^2)(y^2)] = (x^8)+(y^8)
3:27
Prove [(x^4)+(y^4)+√2(x^2)(y^2)][(x^4)+(y^4)-√2(x^2)(y^2)] = (x^8)+(y^8)
21 күн бұрын
Prove (x-y){(x^6)+[(x^5)y]+[(x^4)(y^2)]+[(x^3)(y^3)]+[(x^2)(y^4)]+[x(y^5)]+(y^6)} = (x^7)-(y^7)
4:06
Prove (x-y){(x^6)+[(x^5)y]+[(x^4)(y^2)]+[(x^3)(y^3)]+[(x^2)(y^4)]+[x(y^5)]+(y^6)} = (x^7)-(y^7)
21 күн бұрын
Prove (x+y){(x^6)-[(x^5)y]+[(x^4)(y^2)]-[(x^3)(y^3)]+[(x^2)(y^4)]-[x(y^5)]+(y^6)} = (x^7)+(y^7)
4:04
Prove (x+y){(x^6)-[(x^5)y]+[(x^4)(y^2)]-[(x^3)(y^3)]+[(x^2)(y^4)]-[x(y^5)]+(y^6)} = (x^7)+(y^7)
21 күн бұрын
Prove (x^7)-(y^7) = (x-y){(x^6)+[(x^5)y]+[(x^4)(y^2)]+[(x^3)(y^3)]+[(x^2)(y^4)]+[x(y^5)]+(y^6)}
8:27
Prove (x^7)-(y^7) = (x-y){(x^6)+[(x^5)y]+[(x^4)(y^2)]+[(x^3)(y^3)]+[(x^2)(y^4)]+[x(y^5)]+(y^6)}
21 күн бұрын
Prove (x^7)+(y^7) = (x+y){(x^6)-[(x^5)y]+[(x^4)(y^2)]-[(x^3)(y^3)]+[(x^2)(y^4)]-[x(y^5)]+(y^6)}
8:27
Prove (x^7)+(y^7) = (x+y){(x^6)-[(x^5)y]+[(x^4)(y^2)]-[(x^3)(y^3)]+[(x^2)(y^4)]-[x(y^5)]+(y^6)}
21 күн бұрын
Prove (x-y)(x+y)[(x^2)+xy+(y^2)][(x^2)-xy+(y^2)] = (x^6)-(y^6)
5:30
Prove (x-y)(x+y)[(x^2)+xy+(y^2)][(x^2)-xy+(y^2)] = (x^6)-(y^6)
28 күн бұрын
Prove [(x^3)+(y^3)+xy√(2xy)][(x^3)+(y^3)-xy√(2xy)] = (x^6)+(y^6)
3:35
Prove [(x^3)+(y^3)+xy√(2xy)][(x^3)+(y^3)-xy√(2xy)] = (x^6)+(y^6)
28 күн бұрын
Prove (x^6)-(y^6) = (x-y)(x+y)[(x^2)+xy+(y^2)][(x^2)-xy+(y^2)]
3:20
Prove (x^6)-(y^6) = (x-y)(x+y)[(x^2)+xy+(y^2)][(x^2)-xy+(y^2)]
28 күн бұрын
Prove (x^6)+(y^6) = [(x^3)+(y^3)+xy√(2xy)][(x^3)+(y^3)-xy√(2xy)]
4:16
Prove (x^6)+(y^6) = [(x^3)+(y^3)+xy√(2xy)][(x^3)+(y^3)-xy√(2xy)]
28 күн бұрын
Prove (x^4)-(y^4) = (x-y)(x+y)[(x^2)+(y^2)]
3:15
Prove (x^4)-(y^4) = (x-y)(x+y)[(x^2)+(y^2)]
28 күн бұрын
Prove (x^5)-(y^5) = (x-y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)]
7:18
Prove (x^5)-(y^5) = (x-y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)]
Ай бұрын
Prove (x^5)+(y^5) = (x+y)[(x^4)-(x^3)y+(x^2)(y^2)-x(y^3)+(y^4)]
7:21
Prove (x^5)+(y^5) = (x+y)[(x^4)-(x^3)y+(x^2)(y^2)-x(y^3)+(y^4)]
Ай бұрын
Prove (x^4)+(y^4) = [(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy]
3:38
Prove (x^4)+(y^4) = [(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy]
Ай бұрын
Prove (x^3)-(y^3) = (x-y)[(x^2)+xy+(y^2)]
4:22
Prove (x^3)-(y^3) = (x-y)[(x^2)+xy+(y^2)]
Ай бұрын
Prove (x^3)+(y^3) = (x+y)[(x^2)-xy+(y^2)]
4:29
Prove (x^3)+(y^3) = (x+y)[(x^2)-xy+(y^2)]
Ай бұрын
Prove (x^2)+(y^2) = [x+y+√(2xy)][x+y-√(2xy)]
4:05
Prove (x^2)+(y^2) = [x+y+√(2xy)][x+y-√(2xy)]
Ай бұрын
Prove [x+y+√(2xy)][x+y-√(2xy)] = (x^2)+(y^2)
3:21
Prove [x+y+√(2xy)][x+y-√(2xy)] = (x^2)+(y^2)
Ай бұрын
Prove [(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy] = (x^4)+(y^4)
3:21
Prove [(x^2)+(y^2)+√2xy][(x^2)+(y^2)-√2xy] = (x^4)+(y^4)
Ай бұрын
Prove (x-y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)] = (x^5)-(y^5)
3:22
Prove (x-y)[(x^4)+(x^3)y+(x^2)(y^2)+x(y^3)+(y^4)] = (x^5)-(y^5)
Ай бұрын
Пікірлер
@bhaskarmiriyala8230
@bhaskarmiriyala8230 2 ай бұрын
😊tq for this video
@MharthysChannel
@MharthysChannel 2 ай бұрын
You're welcome. Thank you as well 🙂
@Kemalist188-o2c
@Kemalist188-o2c 2 ай бұрын
Thank you
@MharthysChannel
@MharthysChannel 2 ай бұрын
You're welcome. Thank you as well 🙂
@ararmy6796
@ararmy6796 3 ай бұрын
Thanks
@MharthysChannel
@MharthysChannel 2 ай бұрын
You're welcome. Thank you as well 🙂
@egenestarr1986
@egenestarr1986 3 ай бұрын
.0523 approximate ...............???
@MharthysChannel
@MharthysChannel 2 ай бұрын
Yes, you're correct 🙂 Those are the first four decimal approximation values of sin 3degrees, 0.0523..... If we use a calculator, the decimal approximation of sin 3degrees in 10 decimal places is 0.0523359562.....
@Pranaywele
@Pranaywele 3 ай бұрын
Is costheta common in addition in last step
@MharthysChannel
@MharthysChannel 2 ай бұрын
Hello and good day🙂 Yes, cos(theta) is common in addition in the last step of the solution. To prove this equation, we will employ the sine function’s addition of two angles formula, sin(A+B) = sinAcosB + cosAsinB, (where A is the first angle and B is the second angle), in words, the sine of angle A plus angle B is equal to the sine of angle A times the cosine of angle B plus the cosine of angle A times the sine of angle B, but in this case A = B. Therefore the original formula becomes, sin(A+A) = sinAcosA + cosAsinA = sinAcosA + sinAcosA = 2sinAcosA, or sin(B+B) = sinBcosB + cosBsinB = sinBcosB + sinBcosB = 2sinBcosB. The left-hand side member of this equation is sin(2theta), which is equal to sin(theta + theta), and as we employ the sine function’s addition of two angles formula above where A = B, sin(theta + theta) = [sin(theta)][cos(theta)] + [cos(theta)][sin(theta)] = [sin(theta)][cos(theta)] + [sin(theta)][cos(theta)] = 2[sin(theta)][cos(theta)]. Take note, in Algebra, as we revisit the commutative property of multiplication where, a times b is equal to b times a, or ab = ba. Hence, the term cos(theta)sin(theta) = sin(theta)cos(theta).
@Pranaywele
@Pranaywele 2 ай бұрын
@@MharthysChannel thanks for replying
@MharthysChannel
@MharthysChannel Ай бұрын
@@Pranaywele You're welcome 🙂
@Lala_Deformus
@Lala_Deformus 3 ай бұрын
Thank you, sir🤓👍
@MharthysChannel
@MharthysChannel 3 ай бұрын
Thank you so much 🙂
@varshasingh-yk5fu
@varshasingh-yk5fu 3 ай бұрын
Gajab bro
@MharthysChannel
@MharthysChannel 3 ай бұрын
Thank you so much 🙂
@neuropandian
@neuropandian 3 ай бұрын
thank u so much 😊
@MharthysChannel
@MharthysChannel 3 ай бұрын
You're welcome 🙂 Thank you so much as well🙂
@Game_tech1805
@Game_tech1805 3 ай бұрын
I dont understand that 2sin¤ ?
@MharthysChannel
@MharthysChannel 3 ай бұрын
Good day 🙂 Sin 2theta = sin (1 theta + 1 theta) = sin(theta + theta). According to the sine function’s addition of two angles formula, sin(A+B) = sinAcosB + cosAsinB, (where A is the first angle and B is the second angle), in words, the sine of angle A plus angle B is equal to the sine of angle A times the cosine of angle B plus the cosine of angle A times the sine of angle B, but in this case A = B. Therefore the original formula will become, sin(A+A) = sinAcosA + cosAsinA, or sin(B+B) = sinBcosB + cosBsinB. Now, sin(theta + theta) = sin(theta)cos(theta) + cos(theta)sin(theta), by commutative property of multiplication, a times b is equal to b times a, or ab = ba. Hence, the term cos(theta)sin(theta) = sin(theta)cos(theta). Therefore, sin 2theta = sin(theta + theta) = sin(theta)cos(theta) + sin(theta)cos(theta) = 2 sin(theta)cos(theta), or xy +xy = 2xy, as we add them algebraically. I hope that I answered you correctly 🙂
@파트라슈-o4l
@파트라슈-o4l 5 ай бұрын
thank you very much
@MharthysChannel
@MharthysChannel 5 ай бұрын
You're welcome 🙂 Thank you as well 🙂
@sanjayprasath4850
@sanjayprasath4850 5 ай бұрын
Then how to find cos 18
@MharthysChannel
@MharthysChannel 4 ай бұрын
I did solve the exact value of cos 18 degrees = √[(10+2√5)]/4 by using the Pythagorean Theorem Relation cos theta = √[1 - (sin theta)^2], where theta = 18deg, or cos 18deg = √[1 - (sin 18deg)^2]. Substitute the exact value of sin 18deg = (√5 - 1)/4 to the formula, in which I've already solved and I've included it in this playlist. After that, evaluate and simplify 🙂
@PrithwirajSen-nj6qq
@PrithwirajSen-nj6qq 5 ай бұрын
Cos 24 degree is also the same.
@MharthysChannel
@MharthysChannel 5 ай бұрын
Yes you are correct 🙂 sin66deg = cos24deg = [1+√5+√(30-6√5)]/8. In a right triangle ACB, let A = 66deg and B = 24deg be the two acute angles and angle C = 90deg. Acute angle A and acute angle B are complementary angles. Complementary Angles: A+B = 90deg. Therefore, A = 90deg-B and/or B = 90deg-A. According to Functions of Complementary Angles sinA = sin(90deg-B) = cosB or sin(66deg) = sin(90deg-24deg) = cos(24deg). Also, in a right triangle ACB, 66deg and 24 deg have cofunction relations, it means that sinA = cosB or sin(66deg) = cos24deg, if we let A = 66deg, B = 24deg, and C = 90deg. 🙂
@KuruVFX
@KuruVFX 5 ай бұрын
Thank you for this 🩵
@MharthysChannel
@MharthysChannel 5 ай бұрын
You're welcome. Thank you as well 🙂
@MDmazharulislamabdullah2006
@MDmazharulislamabdullah2006 5 ай бұрын
many many thanks এক জায়গায় প্রমাণ কিন্তু কোথাও পায়নি অবশেষে আপনি সমাধান দিলেন অনেক ধন্যবাদ
@MharthysChannel
@MharthysChannel 5 ай бұрын
Thank you so much 🙂
@abhishekmandhare1323
@abhishekmandhare1323 6 ай бұрын
How to find cos 24 then ??
@MharthysChannel
@MharthysChannel 6 ай бұрын
Hello and good day. To solve for the exact value of cos 24 deg, I've used the cosine function's subtraction formula, that is, cos 24 deg = cos (60 deg - 36 deg) = (cos 60 deg)(cos 36 deg)+(sin 60 deg)(sin 36 deg) = [1+√5+√(30-6√5)]/8. Cos 60 deg = 1/2 and sin 60 deg = (√3)/2. The exact values of cos 60 deg and sin 60 deg can be derived from a right triangle with interior angles: 60 deg, 90 deg, and 30 deg, with corresponding sides: (√3), 2, and 1. Cos 36 deg = (1+√5)/4 can be solved by using the double-angle formula: cos [2(18 deg)] = cos 36 deg = 1-2(sin 18 deg)^2, the exact value of sin 18 deg = (√5-1)/4 can be solved by the aid of a right triangle with interior angles: 36 deg, 90 deg, and 54 deg, while sin 36 deg = [√(10-2√5)]/4 can be solved by using the Pythagorean Theorem Relation: sin 36 deg = √[1-(cos 36 deg)^2]. All of these exact values can be watch in my The Exact Values of sin & cos Functions of a Right Triangle playlist kzbin.info/aero/PL735Z7G5RWRxIDwSoZ2SsWYPn8Mj2-C48 . Thank you so much 🙂
@tyctamil2962
@tyctamil2962 7 ай бұрын
How to find sin18degree
@MharthysChannel
@MharthysChannel 6 ай бұрын
Good day. I did solve it by the aid of a right triangle with interior angles: 36 deg, 90 deg, 54 deg. Through it, we can use the Functions of Complementary Angles Formula: cos⁡𝐴 = ⁡cos⁡(90°⁡−⁡𝐵) = ⁡sin⁡𝐵, in which we can find a quadratic equation. By employing the Quadratic Formula, we can then find the exact value of sin 18 degrees = [(sq. root of 5) -1]/4. I have an existing uploaded video here on my channel about the exact value of sin 18 deg, but I've just recently uploaded a new one. Here's the link just in case you want to watch it: kzbin.info/www/bejne/d6XZc6CLpZJ_p5o . I hope it helps. Thank you so much 🙂
@PinakChhajwani
@PinakChhajwani 7 ай бұрын
Thank you so much, i couldn't understand this before
@MharthysChannel
@MharthysChannel 7 ай бұрын
You're welcome. Thank you so much as well 🙂
@Dojaesd
@Dojaesd 7 ай бұрын
very cool!
@MharthysChannel
@MharthysChannel 7 ай бұрын
Thank you so much 🙂
@ecksdeedeedee
@ecksdeedeedee 8 ай бұрын
nice
@MharthysChannel
@MharthysChannel 7 ай бұрын
Thanks 🙂
@anitamagar51
@anitamagar51 9 ай бұрын
Good 🌟
@MharthysChannel
@MharthysChannel 7 ай бұрын
Thanks 🙂
@mohdistiyak1264
@mohdistiyak1264 Жыл бұрын
Excellent
@MharthysChannel
@MharthysChannel Жыл бұрын
Thank you 🙂
@ChandankumarYadav-wu6ro
@ChandankumarYadav-wu6ro Жыл бұрын
😮
@MharthysChannel
@MharthysChannel Жыл бұрын
Thanks 🙂
@augustosabonete9140
@augustosabonete9140 Жыл бұрын
So good! Thanks!!
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome 🙂 Thank you as well 🙂
@ssjsksll-qb1fz
@ssjsksll-qb1fz Жыл бұрын
thanks a lot
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome.
@ssjsksll-qb1fz
@ssjsksll-qb1fz Жыл бұрын
thanks
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome.
@user2-1
@user2-1 Жыл бұрын
Thanks man
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome.
@its_nilesh_v_k
@its_nilesh_v_k Жыл бұрын
Valuable video thanks sir
@MharthysChannel
@MharthysChannel Жыл бұрын
Thank you so much 🙂
@its_nilesh_v_k
@its_nilesh_v_k Жыл бұрын
Thank you sir for this
@MharthysChannel
@MharthysChannel Жыл бұрын
You're so welcome. Thank you as well.
@АртёмМ-к4л
@АртёмМ-к4л Жыл бұрын
👍🏻
@MharthysChannel
@MharthysChannel Жыл бұрын
Thanks :)
@AshishKumar-d5c9i
@AshishKumar-d5c9i Жыл бұрын
Thank you sir 🎉
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome. Thank you as well.
@Yogesh-747
@Yogesh-747 Жыл бұрын
Thankyou 🙏 sir
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome. Thank you as well.
@claudiocosta4535
@claudiocosta4535 Жыл бұрын
Using cos(15)=cos(45-30) is much easier. Thanks for your video
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome. Thank you as well.
@A-hb5wx
@A-hb5wx Жыл бұрын
Thank you my guy
@MharthysChannel
@MharthysChannel Жыл бұрын
Thank you as well.
@寝-f2g
@寝-f2g Жыл бұрын
Thank you😊
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome :)
@anjinappa919
@anjinappa919 2 жыл бұрын
Thank you sir.... 🙏
@MharthysChannel
@MharthysChannel Жыл бұрын
Thank you as well. You're welcome :)
@pisibell3474
@pisibell3474 2 жыл бұрын
thank you that was a great video
@MharthysChannel
@MharthysChannel Жыл бұрын
You're welcome. Thank you as well :)
@movieday7007
@movieday7007 2 жыл бұрын
It is true sir
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you :)
@marymartin81
@marymartin81 2 жыл бұрын
Galing!
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you so much :)
@jesuschrist.sathish362
@jesuschrist.sathish362 2 жыл бұрын
Thank you bro Useful video for me
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you as well. You're welcome :)
@lseansiamz4251
@lseansiamz4251 2 жыл бұрын
Most precious video
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you :)
@getlosscorona453
@getlosscorona453 Жыл бұрын
Realy
@MharthysChannel
@MharthysChannel Жыл бұрын
Thank you :-)
@sanjayseebaran2201
@sanjayseebaran2201 2 жыл бұрын
amazing
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you :)
@younggoltom1963
@younggoltom1963 2 жыл бұрын
thank u
@MharthysChannel
@MharthysChannel 2 жыл бұрын
You're welcome :)
@doremonfullepisodes6032
@doremonfullepisodes6032 2 жыл бұрын
How is sin 45=root 2/2
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Good day! sin 45deg = √2/2 is derived from the right triangle with sides 1, 1, √2 with internal angles 45deg, 90deg, 45deg. Now, sin 45deg = 1/√2. If you notice, the ratio 1/√2 contains a denominator with a radical. In order to eliminate that radical, we must rationalize the denominator. To do that, insert the factor (√2/√2), multiply it to 1/√2. Then, (1/√2) x (√2/√2) = √2/(√2 x √2) = √2/2. Therefore, sin 45deg = √2/2. Thanks :)
@Aditya___1
@Aditya___1 2 жыл бұрын
Thanks sir
@MharthysChannel
@MharthysChannel 2 жыл бұрын
You're welcome.
@shanmugathasvithushan9109
@shanmugathasvithushan9109 2 жыл бұрын
Thank you so much
@MharthysChannel
@MharthysChannel 2 жыл бұрын
Thank you as well.
@procricket7325
@procricket7325 3 жыл бұрын
thanks man
@MharthysChannel
@MharthysChannel 2 жыл бұрын
You're welcome.
@gamerrathore9971
@gamerrathore9971 3 жыл бұрын
Thks sir
@MharthysChannel
@MharthysChannel 3 жыл бұрын
You're welcome.
@Dipu_.-bhai
@Dipu_.-bhai 3 жыл бұрын
Thank you sir
@MharthysChannel
@MharthysChannel 3 жыл бұрын
You're welcome. Thank you as well.
@pawansinghrawat6941
@pawansinghrawat6941 3 жыл бұрын
Nice
@MharthysChannel
@MharthysChannel 3 жыл бұрын
Thank you.
@mukeshagrawal6924
@mukeshagrawal6924 3 жыл бұрын
thnq sir
@MharthysChannel
@MharthysChannel 3 жыл бұрын
Thank you too.