Trigonometry ke top 10 sawal kaise lage mere bhai😊😊😊 videos ko jyada se jyada share kare , dosto se subscribe bhi karwaye ☺️☺️ Aur kis chapter ke top 10 questions solve karna chahte hai ??? Comments me batao🙂
Q11 option d (4+q2)3/2 sec +tan3a coseca 1/c + sin3/cos3 *1/sin3 1/c + sin2a/cos3a On solving = sec3a And put Value from below Tan2a +1= 3+Q2 +1 Sec2a = 4+q2 (sec2a)3/2 or sec3a = (4+q2)3/2 ans
एक पुल की मेहराब वृत्त की चाप के आकार की है। यदि पुल की चौड़ाई 40 m है और इसके मध्य भाग की ऊँचाई 8 m है, तो पुल की वक्रता त्रिज्या क्या है ? Please solve this question ❓
SecA+ tan ^2A*CosecA=Sec^3A. tan^2A=3+Q^2. Sec^2-1=3+Q^2. Sec^2A=4+Q^2. SecA=(4+Q^2)^1/2. Sec^3A= (4+Q^2)^3/2. Correct answer is :-. D. Thank you so much sir ji 🙏❤️
Option D is the absolute answer. As we see from the given expression, it could be reduced to sec a(1+tan²a), or (1+tan²a)½(1+tan²a)= (1+tan²a)^1.5. Thus, the answer is, (1+3+a²)^1.5= (4+a²)^1.5.
Put Q=0 then a=60° After solution option (d) satisfy the condition. So option (d) is correct I scored 10/10 😊😊 All questions are repeated in practice session on TH-cam by gagan sir🤗🤗
D Es question को पहले long method se करता था but जब से आपका पहले के video me इस question ka solution dekha hai tab se short me hi ho जाता hai Thank you sir ❤
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Trigonometry ke top 10 sawal kaise lage mere bhai😊😊😊 videos ko jyada se jyada share kare , dosto se subscribe bhi karwaye ☺️☺️
Aur kis chapter ke top 10 questions solve karna chahte hai ??? Comments me batao🙂
Bahut hi laa jabab question the guru jii
Algebra ke guru ji
Bhut badhiya solution karwaya sir apne sir please number system ke har ek topic ke bhi top 10 questions Kara do
Mixture allegation ~~ multiplying factor wala
Guruji har chapter k hi kra dijiye concept k sath. Bhot productive h accha revision ho rha hai.
SecA+tan³A.CosecA= SecA(1+tan²A)= Sec³A
▶️tan²A=3+Q²
▶️1+tan²A=4+Q²
▶️Sec²A=4+Q²
▶️SecA= (4+Q²)½
▶️Sec³A= (4+Q²)³/²
Option D ✅
Bahi aapne ye type kaise kiya hai ?
D
Bahut sahi bro
Opt d
Thanks brother 😊
Seca+tan^3a.coseca
Seca+sin^3a/cos^3a.coseca
Seca+tan^2a.seca
Seca(1+tan^2a)
Sec^3a
(√1+tan^2a)^3
(4+Q^2)^3/2
Q.11- 42:10 Ans. (D)❤ chlo ab to sir ke ashirbad se CGL nikal jayga
Q11 option d (4+q2)3/2
sec +tan3a coseca
1/c + sin3/cos3 *1/sin3
1/c + sin2a/cos3a
On solving = sec3a
And put Value from below
Tan2a +1= 3+Q2 +1
Sec2a = 4+q2
(sec2a)3/2 or sec3a = (4+q2)3/2 ans
11.Ans-D (4+Q^2)3/2
1/c3=sec3
Sec2-1 =tan2
Put in eq and solve it
option d is coreect
*_"अतीत पर रोने से बेहतर है नए भविष्य का निर्माण कीजिए.,क्योंकि अतीत एक सबक है भविष्य नहीं..😊......!_*
जय हिन्द गुरुदेव,,❤️🔥🇮🇳
ANS. (D) (4+Q^2)^3/2 JAI JAI SHIYA RAM SIR JI
Tan²a= 3+Q².............................. seca+tan³a×coseca= 1/ cos³a = sec³a
1+tan²a = 1+3 + Q²
Sec² = 4 + Q²
Sec = ( 4+Q²)1/2
Sec³a = (4+Q²)3/4
3/4 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂pagal
(d) (4+Q^2)^3/2
Let ,Tan A = tan 60 = √3
3 = 3 + Q²
Q² = 0
SecA + Tan³A.CosecA =
2 + 3√3× 2/√3 = 8
Ans (d) (4+Q²) = 8 ⚜️
Main jb. Kam time m questions ko slove lar leti huu to apnne ap pe gurv hone lagta hhh......because of sir
Value putting se jldi hoga Q=0
tana=√3.
a=60
Seca+(tan)2.coseca=8
And d is the right option
Jai shree Ram sir ji ❤❤❤
D
Nice class sir ji ...
by the way answer is option D: Thank you.
Putting Q=0
Tana=√3
A=60⁰
Now sec60⁰+tan³60.cosec60⁰
=1/2+3√3×2/√3 =8
Satisfying option 4. At Q=0✅✅✅
Seca +tan3a. Coseca
=Seca(1+tan2a)
=Seca.sec2a
=Sec3a
=(4+Q2)3/2
Let, Q=0,
Then, a=60°
2+3√3•2÷√3
=8
So, option d is correct 💯
😊
Q11.. by value putting
Put Q=0.. so a = 60degree
Thus solving equation by putting 60
We get 8
Only option D makes it 8
Pahle math se dar lagta tha per aab sab se jayada interest hi math me ho Gaya
Batch liya bhai aapne
एक पुल की मेहराब वृत्त की चाप के आकार की है। यदि पुल की चौड़ाई 40 m है और इसके मध्य भाग की ऊँचाई 8 m है, तो पुल की वक्रता त्रिज्या क्या है ? Please solve this question ❓
Let radies is R
Then remaining part is R-8
R^2=20^2+(R-8)^2
Like for Gagan sir ❤❤❤❤
Sir Homework question's answer-------option (D) is right answer (4+Q^2)^3/2
Right answer
@@MathsConceptKing_ 2024 me rank nikalkar aapse Milne aaunga sir 🙏🥰
Sir Mera Selection Hone ke baad Main apne Salary ka 25 ℅ Aapko Donate karunga / Yearly❤
Achcha joke tha bhai
Brother selection only maths se nhi hoga
Put q=0. a=60⁰ ( 2+3root3×2/root 3=8 that is in option D
Kon kon sir ko dil se pasand karta hai ❤❤
Tere andar bhi kaun kaun wala bhut ghus gaya k
Tumera bhi fefda dhadakta hh kya..❤❤
hmare pass dil hi nahi h😂😂tut chuka h😂😂
Hum toh gurde se pasand karte hain dil toh kisi or ko de diya hai n isliye
By value putting
a=60° and Q=0
Correct ans. Option D
SecA+ tan ^2A*CosecA=Sec^3A.
tan^2A=3+Q^2.
Sec^2-1=3+Q^2.
Sec^2A=4+Q^2.
SecA=(4+Q^2)^1/2.
Sec^3A= (4+Q^2)^3/2.
Correct answer is :-. D.
Thank you so much sir ji 🙏❤️
42:03 (d) (4+Q^2)^3/2
Kese aaya
Option D is the absolute answer. As we see from the given expression, it could be reduced to sec a(1+tan²a), or (1+tan²a)½(1+tan²a)= (1+tan²a)^1.5. Thus, the answer is, (1+3+a²)^1.5= (4+a²)^1.5.
OPTN D....(4+Q^2)^3/2 ANSWER.......THANK YOU SIR FOR OUTSTANDING SESSION NAMASTE 🙏🏻 ❤
Right answer
D will be the ans using value putting concept
Put a=60 degree
Question no. 11 , Ans - D
Option D is correct ..... (4+Q²)³/²....
Thanku so mch sir for this amazing class 🥰🤗......
#Mathsguru🤞🏻
Put Q=0 then a=60°
After solution option (d) satisfy the condition. So option (d) is correct
I scored 10/10 😊😊
All questions are repeated in practice session on TH-cam by gagan sir🤗🤗
Yes, correct
@@MathsConceptKing_ Thank you sir I request you please start something new 🆕.
Q.11...Ans-D (4+Q^2)^3/2.
Ans : option D
Let Q=0 then a=60
2+3√3*2/√3=8
Byoption d is ans
Q. No. 11 answer - d - (4+Q²)³/²
Q10..D.( 4+Q^2)^3/2
Ans d hoga after solving given equation we get - sec^3 , we can find out its value using tan^2 =3+Q^2
option D(4+Q²)³/²
(4+Q²)³’² is the right answer sir ❤❤❤❤❤ option (d) is the right answer sir ❤❤❤❤❤❤❤❤❤❤❤
Option --- ddd ((4+Q²) ³/²)
Q11-ans option d from value putting if a=60°
Ans:-Option -D
tan²a=3+Q²
By solving we get;
sec³a=[4+Q²]³/² is right answer 💯❤️ guru jiiii 🙏🙏❤️🤔🤗💪
Option - (D)
D Answer
Q ko 0 man lenge
A=60°
Ans.11 (d) (4+Q²)³/²
41:55 (4+Q²) ³/²
HW-option D. Added 4 both sides of equation
11 Ans option D
option D correct answer of question no, 11 (4+Q^2)^3/2
11..Ans👉👉(4+Q^2)3/2
❣️❣️❣️🙏🙏🙏
Ans D - (4+Q^2)3/2
Option d 41:57
Superb session sir maja aa gya. But end me majboor hokar swal ko solve karna pada (4+Q^2)^3/2 d option answer hoga
Option D is the Right Answer ❤️❤️🙏🙏
((4+Q^2)^3/2 Ans d
D
Es question को पहले long method se करता था but जब से आपका पहले के video me इस question ka solution dekha hai tab se short me hi ho जाता hai
Thank you sir ❤
Ans d (4+Q^2)^3/2
(D)(4+Q^2)^3/2
Ans- D Jai Mahakal Gurudev
answer option D (4+Q^2)^3/2
Q:11 Ka answer (4+Q^2)^3/2
Ans D (4+Q^2)3/2❤
(D). (4+Q^2)^3/2
Ans : Option D
Ans (d).I have solved it by using value putting.😊
Ye series daily chalao sir jee
Very helpful series
Ans .option (d)( 4+Q^2)^3/2
Option -d (4+Q^2)^3/2
D (4+Q^2)^3/2
शब्द कम पड़ जाते है...
आपके मेहनत के बदले शुक्रिया बोलने के लिए।❤❤❤
So nice of you ❤❤
Sir option (D)- (4+Q²)^3/2 is the right answer sir ji 🥰🙏🙏❤️.
Sabse jyada majaa aaya sir is class me. So thank you so much sir ji 🥰🙏🙏❤️
Thanks a lot guru ji & lot of love❤
Ans d
Ans :- D
Option -d : (4+Q²)³/²
Ans- D (4+Q^2)^3/2 , Gjb 💥
Answer - d ❤
mera bhi
11-d ans❤
Ans : D
Option D = (4+Q^2)^3/2
11 ka dd hoga मेरे गुरु देव (1+ tan square) power 3/2 formula से solve हो gya
(4+Q^2)^3/2. Option D . Best solution sir ❤️❤️
Right answer
Q.11- answer (d)
Ans {d}=(4+Q)^3/2
*Q^2
Ans: d
option d (1+Q^2)^3/2
answer no D (4+Q^2)^3/2 🙏🙏
Ans. Option D
Option D is the right answer sir
Thank you for the session sir 🙏
Right answer
D >ans
(4+Q^2) ^3/2
Answer Q=0 then thita 60 then. Put the question ans 8 isiliye option D will be right ans guru ji ❤❤
Right answer
@@MathsConceptKing_ thanks sir
Q)11-D
Put Q=0
Then a =60
Option d ✅
Ans-(d)
sec+tan³cosec= sec³✓
Sec²-1=3+Q²
Sec²=4+Q²
Sec³=(4+Q²)³/²
❤❤❤
Indian ke no.1 teacher
Lekin kya kare batch nahi le paunga
Plz sir help me😭👏👏👏🥰
New zero to hero live batch will be start on 17th March
New zero to hero vod batch available on career will app
6388974650 call or whatsapp on this no any query