Question #97936
(-tan)^2/(1-cot)^2 = tan^2
1
Expert's answer
2019-11-05T12:05:47-0500

Solution:


((tana)/(1cota))2tana2=0((-tan a)/(1-cot a))^2-〖tan a〗^2=0


(tana/(cota1))2tana2=0(tan a/(cot a-1))^2-〖tan a〗^2=0


tana2(1/(cota1)21)=0〖tan a〗^2 (1/(cot a-1)^2 -1)=0


tana2=0or1/(cota1)2=1〖tan a〗^2=0 \,or \, 1/(cot a-1)^2 =1

If tana2=0〖tan a〗^2=0 , then tan a=0, so the angle will be a=πn, n \in Z.

If 1/(cota1)2=11/(cot a-1)^2 =1 , then cot a =2, so the angle will be a=arccot 2+πm, m \in Z

                                 or cot a = 0, but in this situation tan a→∞

Answer:  πn, n \in Z; arccot 2+πm, m \in Z.


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