Find the value of `sin^(-1)(sqrt3/2)+cos^(-1)(-1)+tan^(-1)(-1)`

1.	Find the value of `sin^(-1)(sqrt3/2)+cos^(-1)(-1)+tan^(-1)(-1)`
Answer» Since the principal values of `sin^(-1)x,cos^(-1)xandtan^(-1)x" lie in "[-pi/2,pi/2].[0,pi],(-pi/2,pi/2)` respectively and in these intervals `sin.(pi/3)=sqrt3/2, cospi=-1 tan.(-pi/4)=-1` `rArr sin^(-1)(sqrt3/2)=pi/3, cos^(-1)(-1)=pi,tan^(-1)(-1)=-pi/4` `rArr sin^(-1)(sqrt3/2)+cos^(-1)(-1)+tan^(-1)(-1)` `=pi/3+pi+(-pi/4)=(13pi)/12`

Discussion

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