JEE

Of course, it is extremely possible to learn AI without JEE or NEET. You may also consider B.Sc. or BCA courses with Artificial Intelligence for specialization, which are usually taken on the basis of 12th marks or your result in the exam, to be conducted by some universities. Moreover, very many online sites provide diplomas, certificates, and post-graduate programs in AI and Machine Learning, where there is no specific eligibility and that one does not need such competitive examination like the ones we have in universities. Emphasize on developing a solid background on mathematics and programming.

No, it is unlikely to get top branches of VNIT  like computer Science or electrical branches you may get metallurgy branch maybe from the score of 95.90 in JEE Mains for general unreserved or EWS female student. It doesn't depend on your home state.

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Yes a student can get Nit Suruthkal's EEE branch with a score of 80 percentile in JEE Mains 2025 exam because the students are from home state and their category is St the excepted rank for admission in EEE branch in this college is 30000 to 50000 and if you are St category caste and you are from that state then it is a golden chance for you to getting admission in Nit Suruthkal's 

A 96 percentile in JEE Mains will put you in the general percentage of score between 33000 and 48000. The score is not enough to get an admission in best IITs or most popular faculty members (such as CSE ) in best NITs. But you can easily get into mid-tier NITs (e.g. NIT Agartala, NIT Raipur, NIT SIlchar, NIT Srinagar) in core subjects such as Civil, Mechanical, Chemical, or even Metallurgy. You may even secure a place in newer (e.g. IIIT Manipur, IIIT Bhagalpur, IIIT Dharwad) institutes to other branches such as ECE. This is also dependent on your category and quota of home state.

1. Amity University
2. Bharati Vidyapeeth's College of Engineering
3. Maharaja Agrasen Institute of Technology
4. Maharaja Surajmal Institute of Technology

sin (A + B) = sinA cosB + cosA sinB

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

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

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

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

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

Double Angle Formulas:

sin 2? = 2 sin? cos?

cos 2? = cos²? - sin²? = 2cos²? - 1 = 1 - 2sin²?

tan 2? = 2 tan? / (1 - tan²? ) 

Triple Angle Formulas:

sin 3? = 3 sin? - 4 sin³?

cos 3? = 4 cos³? - 3 cos?