" (a) "(1)/(80)" anfirst ":sin^(-1)x+tan^(-1)x=pi/2

If A = (1)/(pi) [{:(sin^(-1)(pix),, tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),,cot^(-1)((x)/(pi))):}] B = (1)/(pi) [{:(-cos^(-1)(pix),, tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),,-tan^(-1)((x)/(pi))):}] then A - B is equal to

Given that , tan^(-1) ((2x)/(1-x^(2))) = {{:(2 tan^(-1) x"," |x| le 1),(-pi +2 tan^(-1)x","x gt 1),(pi+2 tan^(-1)x"," x lt -1):} sin^(-1)((2x)/(1+x^(2))) ={{:(2 tan^(-1)x","|x|le1),(pi -2 tan^(-1)x","x gt 1 and ),(-(pi+2tan^(-1))","x lt -1):} sin^(-1) x + cos^(-1) x = pi//2 " for " - 1 le x le 1 sin^(-1) ((4x)/(x^(2)+4)) + 2 tan^(-1)( - x/2) is independent of x , then

Given that , tan^(-1) ((2x)/(1-x^(2))) = {{:(2 tan^(-1) x"," |x| le 1),(-pi +2 tan^(-1)x","x gt 1),(pi+2 tan^(-1)x"," x lt -1):} sin^(-1)((2x)/(1+x^(2))) ={{:(2 tan^(-1)x","|x|le1),(pi -2 tan^(-1)x","x gt 1 and ),(-(pi+2tan^(-1))","x lt -1):} sin^(-1) x + cos^(-1) x = pi//2 " for " - 1 le x le 1 If cos^(-1). (6x)/(1 + 9x^(2)) = - pi/2 + 2 tan^(-1) 3x" , then " x in

A : The maximum value of f(x)=Sin^(-1)x+Cos^(-1)x-Tan^(-1)x" is "3pi//4 R : Sin^(-1)x+Cos^(-1)x=pi//2AAx inR

Find the range of f(x)= (1)/(pi)sin^(-1)x+tan^(-1)x+(x+1)/(x^(2)+2x+5)

If A=(1)/(pi)[{:(sin^(-1)(pix),tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),cot^(-1)(pix)):}] and B=(1)/(pi) [{:(-cos^(-1)(pix), tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),-tan^(-1)(pix)):}] then A-B is equal to

Solve for x : i) cos(sin^(-1)x)=1/2 ii) tan^(-1)x=sin^(-1)1/sqrt(2) iii) sin^(-1)x-cos^(-1)x=pi/6

If x ne n pi, x ne (2n+1) (pi)/(2) n in Z, then : sin^(-1) ((sin^(-1) (cos x) + cos ^(-1) (sin x))/(tan^(-1) (cot x) + cot^(-1) (tan x )))=

If A=(1)/(pi)[{:(sin^(-1)(pix),tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),cot^(-1)(pix)):}],B=(1)/(pi)[{:(-cos^(-1)(pix),tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),-tan^(-1)(pix)):}] , then A-B is equal to ......