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tan( 90*1 +30), which lies in the second quadrant and tan is negative in the second quadrant , plus odd multiple of 90 so it becomes -cot(30). Hence tan (90+α)= -cotα. But this works only because we know that α <90.If α>=90 we could just write it as tan ( 90*2+β) and the value would change accordingly.

Keeping this in mind, what if i am with a problem like tan2x= -cot(x+π/3). My goal is to convert the LHS to a tan function. Using tan (90+α)= -cotα (that's what my teacher did) we can write tan2x= tan(π/2+x+π/3) and continue solving. But how can we use the above method without knowing if x >90 or<90? really confused