Guys as i don't take any tutions and my elder sister is busy in her exams so there is no one to teach me some maths problems i face everyday So i wi ask here question and proably expect reply from you guys :)

sure, give us a go, i love a good maths problem....... i have to help my kids daily with their school work !!!!!

This is a trig. identity
(cosA-sinA+1) /(cosA+sinA-1)=cosecA+cotA
LHS
= (cosA-sinA+1) /(cosA+sinA-1)
Multipy the numerator and denominator by (cosA-sinA+1)
(cosA-sinA+1)^2 /[(cosA+sinA-1)(cosA-sinA+1)]
= (cos^2A + sin^2A + 1 - 2sinAcosA + 2cosA - 2sinA) / (cos^2A - sin^2A - 2sinA - 1)
= (2- 2sinAcosA + 2cosA - 2sinA) / (-2sin^2A - 2sinA)
Divide the numerator and denominator by -2
(sinAcosA + sinA - cosA - 1) / (sinA(1+sinA))
= (sinA(cosA+1) + 1(cosA+1))/(sinA(1+sinA))
= (1+sinA)(1+cosA) / (sinA(1+sinA))
= (1+cosA)/sinA
= 1/sinA + cosA/sinA
= cosecA + cotA
= RHS
hope you get it !!!!!!!!

The trick to simplify is to divide all the entities by SinA. The RHS will become(CotA-1+CosecA)/CotA+1-CosecA)=[CotA-(1-CosecA)]/[CotA+(1-CosecA)]Now multiply Nr and Dr by [(CotA)-(1-CosecA)]Now simplify using the identity Cosec(sq)A = 1+ Cot(sq)AYou will get LHS

written in a different way, abit longer but may help
sinA(1+tanA) + cosA(1+cotA)= secA + cosecA
Remember:
tanA = sinA / cosA
cotA = cosA / sinA
secA = 1 / cosA
cosecA = cscA = 1 / sinA
So the identity can be rewritten as:
sinA(1+sinA/cosA) + cosA(1+cosA/sinA) = 1/cosA + 1/sinA
sinA + sin^2A/cosA + cosA + cos^2A/sinA = 1/cosA + 1/sinA
sinA + sin^2A/cosA + cosA + cos^2A/sinA - 1/cosA - 1/sinA = 0
Find the common denominator to add all this stuff up (cosAsinA)
sin^2AcosA/sinAcosA + sin^3A/sinAcosA + sinAcos^2A/sinAcosA + cos^3A/sinAcosA - sinA/sinAcosA - cosA/sinAcosA = 0
(sin^3A + sin^2AcosA - sinA - cosA + sinAcos^2A + cos^3A) / (sinA * cosA) = 0
Now Factor
[(cosA + sinA)(cos^2A + sin^2A - 1)] / (sinA * cosA) = 0
Remember: cos^2A + sin^2A = 1; so:
[(cosA + sinA)(1 - 1)] / (sinA * cosA) = 0
[(cosA + sinA)(0)] / (sinA * cosA) = 0
(0) / (sinA * cosA) = 0
0 = 0
can't think of any other way of explaining it tho

Resitivity = (R X A)/l
where , R - resistance
A - cross section = (π d²)/4
l - length
The answer to your second question lies in the concept of resistivity. As d is directly proportional to Resistivity. Higher the d, higher the resistivity.
And, as Resistance is proportional to resistivity :-


Therefore, thicker the wire ,lower the resistance. Current will easily through thick wire than the thin one.

neah we just get old in these days

isn't it funny once you leave school, all this sh*t means nothing (Except a few rocket physic freaks) as long as i can count my change from a shop or my dealer i couldn't give a sh*t...

Area of a circle is = πr²
As , r = d/2
Therefore ,Area = π X (d/2)² = (π d²)/4

Ha. This was like reading Latin. I'm glad I didn't come here to provide an answer, because it would be something like "...uh...yeah" *walks away*