A particle of mass 1.5kg rests on a rough
plane inclined at 45° to the horizontal.
It is maintained in equilibrium by a
horizontal force of p newtons. Given
that the coefficient of friction between
the particle and the plane is (1/4), calculate
the value of p when the particle is on
the point of moving
i. down the plane
ii. up the plane
[take g=10ms^(−2) ].
If A,B,C are angles of a triangle show that
tan^(−1) (cot Acot B)+tan^(−1) (cot Bcot C)+tan^(−1) (cot Ccot A)
= tan^(−1) {1+((8cos Acos Bcos C)/(sin^2 2A+sin^2 2B+sin^2 2C))}
L′Hopital rule
lim_(x→α) ((f(x))/(g(x)))=lim_(x→α) ((f ′(x))/(g′(x)))= ((f ′(α))/(g′(α)))
f ′(x)=(d/dx)f(x) , g′(x)=(d/dx)g(x) differential
What is it? Proof of the rule.. plz :)