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IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Quadrilateral is simpler. All four sides to choose are predefined. So just choose a point each.
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Rinse and Repeat
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Choose three sides, choose a point each
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
You need to go casewise
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
So it's basically a lot of addition
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Oh yeah nvm
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Quadrilaterals, you can have at most two pairs which are collinear so id say go case wise.
22 replies
IiTeachChem
•Created by Say_miracle_shadow on 1/20/2025 in #💭│doubts
PnC no. of tri<s
Triangle is easy. nC3 - number of ways choosing three collinear points
22 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
SHM
Your I is wrong as far as I can see.
10 replies
IiTeachChem
•Created by Real potato on 1/20/2025 in #💭│doubts
Reimer Tiemann
Carbene intermediate se chalta hai iirc
17 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
Trigonometry
....bhai, cot(A-B) expansion nahi pata hai kya?
35 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
Trigonometry
-3 should be the answer.
35 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
Trigonometry
And the rest cancel out uirc
35 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
Trigonometry
Do it in terms of cot(A+B) for the first two terms (you get 60° and 120°) and cot (A-B) for the last one, as you get 60°
35 replies
IiTeachChem
•Created by flower on 1/20/2025 in #💭│doubts
Trigonometry
I think I did this question once here.
cotAcotB can be written in terms of cot(A±B) and cotB±cotA
35 replies