Does \(\tan\left(-7\right)\) = \(\tan^{-1}\left(7\right)\) and if so, why? Is there a law to this? Will this work to both sin and cos where the negative sign of an answer transfer to a \(\sin^{-1}\) and \(\cos^{-1}\) to a positive answer? Thanks
There's clearly some confusion about notation here. The function \(\tan^{-1}(.)\) (which in VCE is more commonly notated as \(\arctan(.)\)) denotes the
inverse function of \(\tan(.)\). In general, \(\tan(a)\neq\arctan(a)\).
I'm not sure what you mean by your second question. For example, \(-\arcsin(-1)=\dfrac{\pi}{2}>0\) (if this is what you're asking).
However, you should note that inverse circular functions are
not in the Methods course.
Edit: are you trying to refer to odd and even functions perhaps?