Question

Check my working and/or reasoning please? Trig Identities.

Answer 1

Solve the following:

\[\arcsin \frac{ x }{ x-1 } + 2 \arctan \frac{ 1 }{ x+1 } = \pi/2\]

I just want to make sure that I didn't come across the answer "the wrong way", haha. Please bear with, there is a lot of working!

Answer 2

So:

\[\sin a = \frac{ x }{ x-1 } \neq \tan b = \frac{ 1 }{ x+1 } \]

\[a + 2b = \pi/2 = \cos(a + 2b) = 0\]

\[\cos a \cos 2b - \sin a \sin 2b = 0\]

\[\cos a = \sqrt (1 - \sin a ^2) = \sqrt(1 - \frac{ x }{ x-1 } ^2 )= \frac{ \sqrt(1-2x) }{ x-1}\]

to be continued

Answer 3

\[\cos b = \frac{ x+1 }{ \sqrt((x+1)^2 +1)} = \frac{ x+1 }{ \sqrt(x^2+2x+2 }\]

\[\sin b = \frac{ 1 }{\sqrt( x^2+2x+2 )}\]

\[\sin 2b = 2 * \cos b * \sin b = 2 * \frac{ x+1 }{ \sqrt (x^2+2x+2) } * \frac{ 1 }{ \sqrt (x^2+2x+2) } = \frac{ 2x+2 }{ (x^2+2x+2) }\]

\[\cos 2b = \cos b^2 - \sin b^2 = \frac{ x+1 }{ \sqrt (x^2+2x+2) } ^2 - \frac{ 1 }{ \sqrt (x^2+2x+2) }^2 = \frac{ x }{ (x^2+2x+2) }\]

to be continued

Answer 4

\[0 =\frac{ \sqrt 1-2x }{ x-1} * \frac{ x }{ x^2 +2x +2 } - \frac{ 2x+2 }{ x^2 +2x +2 }*\frac{ 1 }{ x^2 +2x +2 }\]

Answer 5

at this point, i have a feeling I need to do more working to "solidly" prove it, but I can see at a glance that x would have to be 0 in order for the end result to be zero. Other than that, have I gone about this the right way? are there some last steps I need to do?

Answer 6

wow lmao