If #cos^-1(x) + cos^-1(y)=pi/2# and #tan^-1(x) +tan^-1(y) =0# then find the value of #x^2+y^2+xy#?

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Answer 1 · 2018-05-09T17:15:21

Here,

#cos^-1 x + cos^-1 y=pi/2...to(1)#

#tan^-1 x +tan^-1 y =0...to(2)#

So,

#tan^-1x=-tan^-1y#

#=>tan^-1x=tan^-1(-y)#

#=>x=-y...to(3)#

#=>x+y=0...to(4)#

Subst. #x=-y# , in #(1)#

#cos^-1 (-y) + cos^-1 y=pi/2#

#pi-cos^-1y+cos^-1y=pi/2#

Then, #cos^-1 x + cos^-1 y=pi/2# does not exists.

Please check your question.
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My suggestion:

If #color(red)(cos^-1 x + cos^-1 y=pi/2 and tan^-1 x -tan^-1 y =0#
,
then find the value of #color(red)(x^2+y^2+xy# .

#cos^-1 x + cos^-1 y=pi/2...to(1)#

#tan^-1 x -tan^-1 y =0...to(2)#

So,

#tan^-1x=tan^-1y#

#=>x=y...to(3)#

Subst, #y=x# in #(1)#

#cos^-1x+cos^-1x=pi/2=>2cos^-1x=pi/2#

#=>cos^-1x=pi/4=>x=1/sqrt2=y#

So, #x^2+y^2+xy=(1/sqrt2)^2+(1/sqrt2)^2+(1/sqrt2)(1/sqrt2)#

#=1/2+1/2+1/2=3/2#

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If my suggestion is true,then please like it.