from inverse trignometric function

sin(sin^-1 1/5 + cos^-1 x) = 1

  Operating both sides by sin^-1 we get (sin^-1 1/5 + cos^-1 x) =sin^-1 1  =/2

Since sin^-1 x + cos^-1 x  = /2

so x =1/5

(ii) tan 1/2 [ sin^-12x/1+x² + cos^-1 1-y²/1+y² ]

tan 1/2[2 tan-¹  x +2tan-¹  y]= tan[tan^-1 x+tan-¹ y]   = tan[tan^-1 (x+y)/(1-xy)]=x+y/(1-xy)

 Similarly try other part.Also in future pls raise only one question in a query