IIT JEE , AIEEE Forum - Mathematics , Trignometry

iitaspirant10

what is the maximum and minimum values of

sin^-1x+cos^-1x+tan^-1x??

aman23iit

hi friend its simple put

sin^-1x+cos^-1x+tan^-1x=f(x)

then you know that

sin^-1x+cos^-1x=pi/2............property

and you also know that the range of tan^-1x is (-pi/2,pi/2)

hence the maximum value of this expression is

pi ...........................where pi is

bye

plzz rate me

shakirshafi12

f(x)=sin^-1x+cos^-1x+tan^-1x

now here the domain of this function will be [-1,1]

f(x)=pi/2+tan^-1x

now for max value x=1

f(x)=pi/2+pi/4=3pi/4

and for minimum value x=-1

f(x)=pi/2-pi/4=pi/4

-------------------------

i think its right!!

if not (mabbe i was in hurry)

newayz!!
keep rocking

iam_quantized

ya shakir i agree with u

avdesh100

i suppose answer is pie

magiclko

here the domain of function is [-1,1] as sin inverse and cos inverse are only defined for theses value of x only....
and for ever |x|<=1,
thrfore f(x) reduces to
f(x)=pi/2+tan^-1x
now as tan inverse is an increasing function in given domain, so for maximum value of f(x),
x=1
=> f(x)= pi/2+pi/4

=3pi/4
and for minimum value
x= -1
=> f(x)=pi/2-pi/4

=pi/4

so shakir is absolutely correct .......

iitaspirant10

aman....ur ans is wrong....voted by mistake

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