urgent ..

formula for sin^-1X + sin^-1Y , sin^-1X -sin^-1Y

cos^-1X + cos^-1Y

cos^-1X - cos^-1Y

prove: tan inverse [ cos x / (1+sin x)] = pie /4 - (x/2)

show that tan(1/2 sin-1 3/4) = 4-root7/3

If cos^-13/5- sin^-14/5 = cos^-1x then x =

a.0 b.1

c.-1 d.2

prove that :

tan [pi/4 + 1/2 cos^-1 a/b] + tan[pi/4 - 1/2cos^-1 a/b] = 2b/a

find the value of x : sin^-1x + sin^-1(1-x) = cos^-1x

Show that tan(1/2sin^-13/4) = (4-Squareroot of 7)/ 3

i have 2 question -

1)if cos_^-1(x/a)+cos^-1(y/b)=d

prove that : x²/a²-2xy/ab-cosd+y²/b²=sin²d

2)solve tan(cos^-1x)=sin(tan^-1x)

3)tan^-1x+tan^-1y+tan^-1z=pi/2

prove that xy+yz+zx =1

2tan^-1(1/3) +cot^-14 =tan^-1(16/13)

cos^-1(x) + sin^-1(x/2)=pi/6

find x.

formula of 1-sinx

Prove that :

tan^-1(1/4) + tan-1(2/9) = 1/2 Cos^-1 (3/5)

sin[ tan^-1 x (1-x^2/2x) + + cos^-1 (1-x^2/1+x^2)]=1pls prove this

if cos^-1x + cos^-1y + cos^-1z = pi

prove that x² + y² + z² +2xyz = 1

find the greatest and least values:(sin^-1x)²+(cos^-1x)²

Write the function, cot^-1 (sqrt(1+x^2)+x) in the simplest form

prove that tan ^-1(1) + tan^-1(2) + tan^-1(3) = pie.

if sin^-1x+sin^-1y=pi/2 then find cos^-1x+cos^-1y

sin^-1(1-x)-2sin^-1x=pi/2

If y=cot^-1(square root cosx) - tan^-1(square root cosx)

prove that sin y= tan² x/2.

if tan-1x+tan-1y+tan-1z=π/2 show that xy+yz+zx=1?

If tan-1{root (1+x²) - root(1-x²)}/{ root (1+x²) +root (1-x²)} =theta , then prove that x² =sin 2 theta .

Solve for x : sin-1 6x + sin-1 6root3 x = -pi/2

solve for X

sin-1x + sin-12x = ∏/3

Solve for the equation cos (tan^-1x) = sin(cot^-13/4)

IF y = sin^-1 (2^x+1/1+4^x) find dy/dx

if sin ^-1x +sin^-1 y+ sin^-1 z= pie ,prove that

a)x( root 1-x²)+y( root 1- y²)+ z(root 1-z²)= 2xyz

b)x⁴+ y⁴+z⁴+ 4 x²y²z²= 2( x²y²+y²z²+ z²x²)

prove that sec square (tan inverse 2) +cosec square (cot inverse 3) =15

Solve for x - tan^-1 (x+1) + tan^-1 (x-1) = tan^-1 8/31

prove that

2tan^-1 [ { (a-b) / (a+b) }^1/2 tan x/2 ] = cos^-1{ (b +acos x ) / (a+bcos x) }

sin^-1 x - cos^-1 x = pi/6

how to solve tan^-1 2x/1-x² + cot^-1 1-x²/2x = pie/3

sin inverse sin 3pi/5????

Please explain me in detail that how [cos inverse(cos 7pi/6)] is equal to 5pi/6.

It is Q. no. 19 of Ex. 2.2 of chapter INVERSE TRIGONOMETRIC FUNCTION (N.C.E.R.T).

If tan^-1x+tan^-1y+tan^-1z=pie, prove x+y+z=xyz

tan^-1 x + 2cot^-1 x = 2pi/3 find value of x

sin(1/2 cos inverse 4/5) ??evaluate

Sin^-1x+Sin^-1y+Sin^-1z=3(pi)/2 then find x+y+z=?

prove that 2 tan-1 (root of a+b/a-b tn theta/2) = cos -1 (acos theta b/a b cos theta)

if two angles of a triangle are tan^-1(2) and tan^-1(3) find the third angle

simplify

sin-1(sinx+cosx/root 2)where pie

Find x if sin^-1(5/x) + sin^-1(12/x) =pi/2

cot[pi/4 - 2cot^-1 3] =7

If sin x₁+sinx₂+sinx₃?? = 3, then write the value of cosx₁+cosx₂+cosx₃??

prove that

(i) sin^-1 (12/13) + cos^-1 (4/5) +tan^-1 (63/16)=0

(ii) 2sin^-1(3/5) - tan^-1 (17/31) =pi / 4

solve:

cos^-1 ([x²-1]/[x²+1]) + tan^-1[(2x)/(x²-1)]=2pi/3

(cosx - cosy)² + (sinx - siny)² = 4 sin²(x-y)/2.

solve for x
cos(2sin^-1x) =1/9 ,x>0