Q:evaluate - i) tan^-1(sin (-pi/2)

ii) tan^-1(2cos (2sin^-1 1/2)iii) sin^-1(2a / (1+a^2) + sin^-1(2b/(1+b^2)=2tan^-1 x

(i)
tan-1(sin(-π2))=tan-1-1=-π4
(ii)
tan-12cos2sin-112=tan-12cossin-12*121-14   [since 2sin-1x=sin-12x1-x2=tan-12cossin-134=tan-12cossin-132=tan-12cosπ3=tan-12*12=tan-1(1)=π4
(iii)
the given equation is:
sin-12a1+a2+sin-12b1+b2=2tan-1xx=tan12sin-12a1+a2+sin-12b1+b2
let a=tanα   and b=tanβ 
2a1+a2=2tanα1+tan2α=2.sinαcosα1+sin2αcos2α=2sinαcosα*cos2αsin2α+cos2α2a1+a2=2sinαcosα=sin2αsin-12a1+a2=sin-1(sin2α)sin-12a1+a2=2α
similarly sin-12b1+b2=2β
therefore
x=tan12*2α+2β
x=tanα+β=tantan-1a+tan-1b=tantan-1a+b1-abx=a+b1-ab

hope this helps you

 

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