Question

Use theta equals inverse tangent of 28/x minus inverse tangent of 6/x to show that theta also equals 22x/(x^2=168).

Answer 1

I will write inverse tangent as arctan
let A = arctan (28/x) and B = arctan (6/x)
and tanA = 28/x and tanB = 6/x

then Ø = A - B
take tan of both sides
tan Ø = tan(A-B)
= (tanA - tanB)/(1 + tanAtanB)
= (28/x - 6/x)/(1 + (28/x)(6/x) )
= (22/x) / ((x^2 + 168)/x^2 )

= 22x/(x^2 + 168)

so tanØ =22x/(x^2 + 168)
then
Ø = arctan (22x/(x^2+168))
or
Ø = inverse tangent (22x/(x^2+168)

I see two typos in your last line