Trigonometry Top-10 सवाल🔥 एक सवाल पक्का Gagan Pratap Sir
Вставка
- Опубліковано 12 бер 2024
- Buy Book:
CGL PRACTICE KING- amzn.to/40qLGVh
CHSL PRACTICE KING- amzn.to/41itIEA
Math Concept King Book- amzn.to/3Ss3FIF
CGL CHSL BOOK- amzn.to/3G4cO33
CPO PRACTICE KING- amzn.to/3QstgP8
Advance Math Book- amzn.to/3FIbjXW
Arithmetic Class Notes- amzn.to/46ZgZZu
Maths 8000+ TCS- MCQ- amzn.to/3s9G6de
❤️। Gagan Pratap Sir
Join Our New Maths Special (ZERO TO HERO BATCH)
Download App- bit.ly/CareerwillApp
Batch Related Query : 6388974650
Telegram Channel :
t.me/mathsbygaganpratap
Maths Concept King (Gagan Pratap Sir)
#maths
#ssc #cgl #gaganpratapmaths
#cpo #gagan_pratap
#chsl #gagan_sir #mathsbygaganpratap
#mathswizardgaganpratapsir
#maths_tricks #profitandloss
#ssc_cgl #commission
#maths_wizard #ratio
#gaganpratapsir #surds
#mathsconceptking #time
#maths_concept_king #surds_and_indices
#geometry #circular_track
#trigonometry #discount
#numbersystem #ratio #ratio_gagan_sir #mensuration #algebra
#algebra #time #speed #train
#simplification #semicircle #surds_and_indices
#ssccgl2023 #chsl_pre #chsl_exam #chsl_pre
#cgl2023 #advance #chsl_2023pre
#cpo #ssc_chsl #chsl_2023 #inradius #linear_race
#maths_concept_king #cgl_reviesed_result #cgl_repeat_questions
#ssc #mains #cgl2023 #cgl_mains
#lcm #maths_concept
Like, Share, Subscribe ❤️
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 😊
*_"अतीत पर रोने से बेहतर है नए भविष्य का निर्माण कीजिए.,क्योंकि अतीत एक सबक है भविष्य नहीं..😊......!_*
जय हिन्द गुरुदेव,,❤️🔥🇮🇳
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
Q.11- 42:10 Ans. (D)❤ chlo ab to sir ke ashirbad se CGL nikal jayga
1/c3=sec3
Sec2-1 =tan2
Put in eq and solve it
option d is coreect
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
ANS. (D) (4+Q^2)^3/2 JAI JAI SHIYA RAM SIR JI
(d) (4+Q^2)^3/2
Mind blowing question❤❤
Super Sir
Main jb. Kam time m questions ko slove lar leti huu to apnne ap pe gurv hone lagta hhh......because of sir
Jai shree Ram sir ji ❤❤❤
D
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. a=60⁰ ( 2+3root3×2/root 3=8 that is in option D
Putting Q=0
Tana=√3
A=60⁰
Now sec60⁰+tan³60.cosec60⁰
=1/2+3√3×2/√3 =8
Satisfying option 4. At Q=0✅✅✅
Like for Gagan sir ❤❤❤❤
Truly amazing class ❤ we need more classes like this 😊😊😊😊 always thank ful of you sir😊😊😊🎉❤❤😂😂😂
Question no. 11 , Ans - D
Pahle math se dar lagta tha per aab sab se jayada interest hi math me ho Gaya
Batch liya bhai aapne
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
Nice class sir ji ...
by the way answer is option D: Thank you.
Option D is the Right Answer ❤️❤️🙏🙏
Let, Q=0,
Then, a=60°
2+3√3•2÷√3
=8
So, option d is correct 💯
😊
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
D will be the ans using value putting concept
Put a=60 degree
Q. No. 11 answer - d - (4+Q²)³/²
Seca +tan3a. Coseca
=Seca(1+tan2a)
=Seca.sec2a
=Sec3a
=(4+Q2)3/2
Ans : option D
Option - (D)
Concept lajwab tha sir ji.....❤❤❤❤❤
option D(4+Q²)³/²
Easy concept ❤❤❤
Ye series daily chalao sir jee
Very helpful series
Option -d : (4+Q²)³/²
Behtareen sir
Thanks a lot guru ji & lot of love❤
Ans d
First try to make tan²a into sec²a then then on solving we tan²a.seca hence the answer is option (d)
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
Bahut majedar hey sir questions
By value putting
a=60° and Q=0
Correct ans. Option D
Ans D - (4+Q^2)3/2
Option -d (4+Q^2)^3/2
(D). (4+Q^2)^3/2
Majedaar sir ji 🥰🥰
Q11-ans option d from value putting if a=60°
Super sir ❤❤
Wonderful session sir❤❤
Keep watching
Option . D
Sir amazing session 🎉❤ Sir VOD agar abhi lete hain toh usme latest questions hongey for cgl 2024?
Thanks so much sir ❤❤
Ans d hoga after solving given equation we get - sec^3 , we can find out its value using tan^2 =3+Q^2
Ans. D (4+Q^2)^3/2
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 🙏❤️
Sir bahut shandar
(4+Q^2)^3/2 option D
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
Q11--option D
Q.11...Ans-D (4+Q^2)^3/2.
answer option D (4+Q^2)^3/2
Option D { Sec³A=( 4+Q²)³/2}
Wonderful session ❤
Glad you enjoyed it!
Ans (d).I have solved it by using value putting.😊
Let Q=0 then a=60
2+3√3*2/√3=8
Byoption d is 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 ❓
Let radies is R
Then remaining part is R-8
R^2=20^2+(R-8)^2
option D correct answer of question no, 11 (4+Q^2)^3/2
Ans - D
d option (4+Q^2)^3/2
Nice session sir thank you so much
Always welcome
*"respect button for Gagan sir,,❤❤😊.....!!*
Ans .option (d)( 4+Q^2)^3/2
D. (4+Q²)3/2
good morning sir , ans (4+Q^2)^3/2
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 🙏🥰
D >ans
ans. - D
Ans. Option D
Ans :- D
Ans : Option 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 ❤
Sir, after reading math from you, I am solving the questions very soon, thank you so much sir.❤
Most welcome
Most welcome
Tnq gurudev 😊
Option --- ddd ((4+Q²) ³/²)
OPTN D....(4+Q^2)^3/2 ANSWER.......THANK YOU SIR FOR OUTSTANDING SESSION NAMASTE 🙏🏻 ❤
Right answer
(4+Q^2) ^3/2
Ans-(d)
Q)11-D
Great sir ji
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 ⚜️
Amazing sir ji
Keep watching
Q11 -ansD
D.(4+Q)^3/2
Gjb 👍🏼
d) (4+Q^2)^3/2
Ans- D (4+Q^2)^3/2 , Gjb 💥
Option D is correct ..... (4+Q²)³/²....
Thanku so mch sir for this amazing class 🥰🤗......
#Mathsguru🤞🏻
Question 11-- option d
Speechless...
Ans option D
Ques 11. Option D