Question

please help with this itf

Answer 1

i dont know because i am in science and english only

Answer 2

tan^-1x=A
x=tanA
√sec²A-1=x
sec²A-1=x²
sec²A=1+x²
secA=√1+x²
cosA=1/√1+x²
A=cos^-1(1/√1+x²)=tan^-1x
sin2[cot^-1{cos(tan^-1x)}]
=sin2[cot^-1{cos(cos^-1(1/√1+x²)}]
=sin2[cot^-1{1/√1+x²}]=sin2B...(1)

let cot^-1(1/√1+x²)=B
1/√1+x²=cotB
cosec²B-1=(1/√1+x²)²
cosec²B=1/(1+x²)+1
=(1+1+x²)/(1+x²)
cosecB=under root(2+x²)/(1+x²)
sinB=√(1+x²)/(2+x²)
si
B=sin^-1[√(1+x²)/(2+x²)]
put in (1)
sinB=sin(sin^-1[√(1+x²)/(2+x²)]
=√(1+x²)/(2+x²)