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urgent ..

formula for sin

^{-1}X + sin^{-1}Y , sin^{-1}X -sin^{-1}Ycos

^{-1}X + cos^{-1}Ycos

^{-1}X - cos^{-1}Yformula of 1-sinx

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

Taking the moon's distance from the earth as 360000 km and the angle subtended by the moon at any point O on earth as half a degree,estimate the diameter of the moon . (Use pie = 3.1416)

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

prove that :

tan [pi/4 + 1/2 cos

^{-1}a/b] + tan[pi/4 - 1/2cos^{-1}a/b] = 2b/afind the value of x : sin

^{-1}x + sin^{-1}(1-x) = cos^{-1}x^{-1}x-cos^{-1}y=cos^{-1}{xy+whole root over(1-x^{2)}(1-y^{2})^{}}Show that tan(1/2sin

^{-1}3/4) = (4-Squareroot of 7)/ 3i have 2 question -

1)if cos

_{-1}(x/a)+cos^{-1}(y/b)=dprove that : x

^{2}/a^{2}-2xy/ab-cosd+y^{2}/b^{2}=sin^{2}d2)solve tan(cos

^{-1}x)=sin(tan^{-1}x)3)tan

^{-1}x+tan^{-1}y+tan^{-1}z=pi/2prove that xy+yz+zx =1

URGENT!!

Show that

sin(4/5) +^{-1}cos(2/square root of 5)=^{-1}cot(2/11)^{-1}Please ans in a simple way

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

find x.

integrate 0-pi x dx/4-cos

^{2x}Prove that :

tan

^{-1}(1/4) + tan-1(2/9) = 1/2 Cos^{-1}(3/5)^{-1}x=tan^{-1}(3x-x^{3}/1-3x^{2})if cos

^{-1}x + cos^{-1}y + cos^{-1}z = piprove that x

^{2}+ y^{2}+ z^{2}+2xyz = 1find the greatest and least values:(sin

^{-1}x)^{2}+(cos^{-}^{1}x)^{2}A. 84448

B. 84450

C. 84449

D. 84077

prove that tan

^{-1}(1) + tan^{-1}(2) + tan^{-1}(3) = pie.^{r}) *^{9}C_{r}) = (alpha * (3/2)^{9}) + beta , then find the value of alpha , beta and alpha + beta . No linksevaluate tan(tan-1 (-4).

cot

^{-1}((ab+1)/(a-b)) + cot^{-1}((bc+1)/(b-c)) + cot^{-1}((ca+1)/(c-a))=pisin

^{-1}(1-x)-2sin^{-1}x=pi/2^{-1}( x+ square root 1-x^{2}/ root 2 in the simplest formIf y=cot

^{-1}(square root cosx) - tan^{-1}(square root cosx)prove that sin y= tan

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

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

2

tan(1/3) +^{-1}cot4 =^{-1}tan(16/13)^{-1}solve for X

sin-1x + sin-12x = ∏/3

Solve for the equation cos (tan

^{-1}x) = sin(cot^{-1}3/4)^{-1 }(x + 1) ) = cos ( tan^{-1}x ) ...... find x ..... i asked this question and an expert proved it wrong .....but this question is absolutely correct ...it's 2015 board question .... so kindly find x ...since question is correct ..!!! @ PRIYANKA KEDIA ......AND OTHER MERITNATION EXPERTS ....!!!if sin

^{-1}x +sin^{-1}y+ sin^{-1}z= pie ,prove thata)x( root 1-x

^{2})+y( root 1- y^{2})+ z(root 1-z^{2})= 2xyzb)x

^{4}+ y^{4}+z^{4}+ 4 x^{2}y^{2}z^{2}= 2( x^{2}y^{2}+y^{2}z^{2}+ z^{2}x^{2})a) 1+x b)1-x c) x d)1/x

If tan-1{root (1+x

^{2}) - root(1-x^{2})}/{ root (1+x^{2}) +root (1-x^{2})} =theta , then prove that x^{2}=sin 2 theta .^{-1 }(cosec π/6) + tan^{-1 }(tan 7π/6)Solve for x - tan^-1 (x+1) + tan^-1 (x-1) = tan^-1 8/31

^{-1}(2-x/2+x) = 1/2 tan^{-1}x/2prove that

2tan

^{-1}[ { (a-b) / (a+b) }^{1/2 }tan x/2 ] = cos^{-1}{ (b +acos x ) / (a+bcos x) }prove that sec square (tan inverse 2) +cosec square (cot inverse 3) =15

^{2}=4x and x^{2}= -32 y .tan

^{-1}( x-1 ) + tan^{-1}x + tan^{-1}( x+1 ) = tan^{-1}3xpls answer , I have a test tomorrow :/

prove that 1/2 tan

^{ -1}x = cos -1 ( 1 + root ( 1 + x^{2})) / 2 root (1 + x^{2}) )^{1/2}how to solve tan

^{-1}2x/1-x^{2}+ cot^{-1}1-x^{2}/2x = pie/3^{-1}[1+tanx/1-tanx] were x related to [-pie/4,pie/4]sin inverse sin 3pi/5????

Find x if tan

^{-1}4 + cot^{-1}x = pi / 2.Write the function, cot^-1 (sqrt(1+x^2)+x) in the simplest form

If tan

^{-1}x+tan^{-1}y+tan^{-1}z=pie, prove x+y+z=xyzSolve the equation:

sin-1 (3x/5) + sin-1(4x/5)= sin-1x

tan

^{-1}x + 2cot^{-1}x = 2pi/3 find value of xcos

^{-1}(cos 5pie/3)sin(1/2 cos inverse 4/5) ??evaluate

^{-1}( 3/5 cos x + 4/5 sin x )prove that 2 tan-1 (root of a+b/a-b tn theta/2) = cos -1 (acos theta b/a b cos theta)

^{-1}( a+x)/a + tan^{-1 }( a-x)/a= pi/6, then prove that x^{2}= 2 root3 a^{2}simplify

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

if sin

^{-1}(1/3) + sin^{-1}(2/3) = sin^{-1}(x). Then x =.... ?Find x if sin

^{-1}(5/x) + sin^{-1}(12/x) =pi/2F(x) is: (where {.} denotes fractional part function and [..] denotes greatest integer function and sgn (x) is a signum function)(A) periodic with fundamental period 1

(B) even

(C) range is singleton

(D) identical to sgn

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

Maximum and minimum values of sin

^{6}x+cos^{6}x isIF y = sin

^{-1}(2^{x+1}/1+4^{x}) find dy/dx^{-1 }(1/2)}cot[pi/4 - 2cot

^{-1}3] =7prove 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

cos cos

^{-1}[-root3/2] +pi/4]solve:

cos

^{-1 }([x^{2}-1]/[x^{2}+1]) + tan^{-1}[(2x)/(x^{2}-1)]=2pi/3^{-1}(1/x)=cot^{-1}x holds^{-1}root(x(x+1)) + sin^{-1}root(x^{2}+x+1) = pi/2.(cosx - cosy)

^{2}+ (sinx - siny)^{2}= 4 sin^{2}(x-y)/2.cos

^{-1}[ 3/5 cosx +4/5sinx ] find its value?I can ' t understand this question . please explain it briefly ?

Simplify

sin ( 2cos

^{-1}x)