161 - Mathematics Questions Answers

Please submit what we have to do and also correct the question 

it is cos3 or cos3x 

I dont know what the question is but if the question is to solve this eq. then the method will be as given below

tan²x tan²3x tan4x = tan²x  - tan²3x + tan4x

so tan4x =  [tan²x  - tan²3x] / [tan²x tan²3x-1]

now split RHS terms by formula x²-y²= (x+y)(x-y)

so RHS = tan4xtan2x 

now solve

by diagram (6,4),(6,3),(6,2) are points having AB<4 besides it two upside points are also there by symmetry not shown in diagram. These points are (6,5) and (6,6)

Total no. of points = 5

It is the theorem for the equilibrium of a body under 3 forces

According to the theorem if three forces are acting on a body and body is in equilibrium then 

P/sinA = Q/sinB = R/sinC

let us consider the case of ellipse with x and y axes as their axes

eq. is x²/a² + y²/b²= 1

on differentiating we get 2x/a² + 2yy^'/b² = 0                                y'is first differential

so yy^'/x = -b²/a²

again differentiate and get answer. 

You should remember that you should differentiate as many times as the number of constants.

for ex. in the case of parabola only first diffrentiation is sufficient.

now complete it for all conics

first use C1 = C1+C2+C3

then R1 = R1+R2 and R2 = R2+R3

after it solve mathematically

x=rcosθ and y=rsinθ

so dx=-rsinθdθ  and dy = rcosθdθ

so p = -cotθ

so given eq. will become x-cotθy = (x-y)cosecθ 

on putting values of x and y we get 

0 = r(cosθ-sinθ)/sinθ

so tanθ = 1  so θ = nπ+π/4

so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)

y²= 4a(x+a)

so 2yy^' = 4a so a = yy^'/2

on putting a we get y²= 4yy^'/2(x+yy^'/2)      

so y² = yy^' (2x+yy^')  or y = 2xy^' + yy^'2    (1)

now on putting -1/y^' in the place of y^'

we get y² = -y/y^'[2x-y/y^']

so -yy^'2 = 2xy^' - y (2)

similarity of (1) and (2) shows that the given curve is self orthogonal 

total numbers = 4*3*2*1 = 24

on arranging these numbers in order we get 1234

so rank = [24/4]*2+[6/3]*1+[2/2]*1+1 = 16

let the 4 digit number is 1000a+100b+10c+d

then according to the given condition b=3a, c=a+b, d=2b

so c=4a, d=6a and b=3a

on putting all the values we get number = 1000a+100(3a)+10(4a)+6a = 1346a

on putting a =1, number = 1346