Inverse Trig Question - Year 11 Maths Ext 1 (1 Viewer)

[Scrambled]

Could I have some help solving this pleaseeee
The answer is pi/8 apparently

 

[WeiWeiMan]

Could I have some help solving this pleaseeee
The answer is pi/8 apparently
View attachment 43352

tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4
tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4
tan^-1(√2-1)+tan^-1(√2-1)=π/4
2tan^-1(√2-1)=π/4
tan^-1(√2-1)=π/8

 

[Scrambled]

tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4
tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4
tan^-1(√2-1)+tan^-1(√2-1)=π/4
2tan^-1(√2-1)=π/4
tan^-1(√2-1)=π/8

ohhhh so there was a typo it was meant to be arctan dang.
Anyways thank you!