Question

tan inverse [2sin(2cos inverse root 3/2)]

Answer 1

tan−1(sin(−π2))=tan−1(−1)=−π4
(ii)
tan−1(2cos(2sin−112))=tan−1(2cos{sin−1(2*121−14‾‾‾‾‾‾√)})   [since 2sin−1x=sin−1(2x1−x2‾‾‾‾‾‾√)=tan−1(2cos{sin−134‾‾√})=tan−1(2cos(sin−13√2))=tan−1(2cos(π3))=tan−1(2*12)=tan−1(1)=π4
(iii)
the given equation is:
sin−1(2a1+a2)+sin−1(2b1+b2)=2tan−1xx=tan[12{sin−1(2a1+a2)+sin−1(2b1+b2)}]
let a=tanα   and b=tanβ 
2a1+a2=2tanα1+tan2α=2.sinαcosα1+sin2αcos2α=2sinαcosα*cos2αsin2α+cos2α2a1+a2=2sinαcosα=sin2αsin−1(2a1+a2)=sin−1(sin2α)sin−1(2a1+a2)=2α
similarly sin−1(2b1+b2)=2β
therefore
x=tan[12*(2α+2β)]
x=tan(α+β)=tan[tan−1a+tan−1b]=tan[tan−1a+b1−ab]x=a+b1−ab

hope this helps you
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