HSC Arts (English Medium) १२ वीं कक्षा - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Show that `cot^-1  1/3 - tan^-1  1/3 = cot^-1  3/4`.

Show that `tan^-1  1/2 = 1/3  tan^-1  11/2`

Show that `cos^-1  sqrt3/2 + 2 sin^-1  sqrt3/2 = (5pi)/6`.

Show that `2 cot^(-1)  3/2 +  sec^(-1)  13/12 = π/2`

Prove the following:

`cos^-1 "x" = tan^-1 (sqrt(1 - "x"^2)/"x")`, if x > 0

Prove the following:

`cos^-1 "x" = pi + tan^-1 (sqrt(1 - "x"^2)/"x")`, if x < 0

If | x | < 1, then prove that

`2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2)) = sin^-1 ("2x"/(1 + "x"^2)) = cos^-1 ((1 - "x"^2)/(1 + "x"^2))`

If x, y, z are positive, then prove that

`tan^-1 (("x - y")/(1 + "xy")) + tan^-1 (("y - z")/(1 + "yz")) + tan^-1 (("z - x")/(1 + "zx")) = 0`

If `tan^-1 "x" + tan^-1 "y" + tan^-1 "z" = pi/2,` then show that xy + yz + zx = 1

If cos^-1 x + cos^-1y + cos^-1z = 3π, then show that x² + y² + z² + 2xyz = 1.

Determine the order and degree of the following differential equation:

`("d"^2"y")/"dx"^2 + "x"("dy"/"dx")` + y = 2 sin x

Determine the order and degree of the following differential equation:

`root(3)(1 +("dy"/"dx")^2) = ("d"^2"y")/"dx"^2`

Determine the order and degree of the following differential equation:

`(dy)/(dx) = (2sin x + 3)/(dy/dx)`

Determine the order and degree of the following differential equation:

`("d"^2"y")/"dx"^2 + "dy"/"dx" + "x" = sqrt(1 + ("d"^3"y")/"dx"^3)`

Determine the order and degree of the following differential equation:

`("d"^2"y")/"dx"^2 + ("dy"/"dx")^2 + 7"x" + 5 = 0`

Determine the order and degree of the following differential equation:

(y''')² + 3y'' + 3xy' + 5y = 0

Determine the order and degree of the following differential equation:

`(("d"^2"y")/"dx"^2)^2 + cos ("dy"/"dx") = 0`

Determine the order and degree of the following differential equation:

`[1 + (dy/dx)^2]^(3/2) = 8(d^2y)/dx^2`

Determine the order and degree of the following differential equation:

`(("d"^3"y")/"dx"^3)^(1/2) - ("dy"/"dx")^(1/3) = 20`

Determine the order and degree of the following differential equation:

`"x" + ("d"^2"y")/"dx"^2 = sqrt(1 + (("d"^2"y")/"dx"^2)^2)`