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I was reading a trig book in which the author had mentioned a trick to jump from $(\theta+90^\circ)$ to $(\theta-90^\circ)$ formulas. I understood it for Sine and Cosine but couldn't for Tangent.

Here is the $(\theta+90^\circ)$ formula…

$$\tan(\theta+90^\circ)=-\cot\theta$$

He mentioned to replace $\theta$ in the above formula by $(\theta-90^\circ$ to obtain the desired $(\theta-90^\circ)$ formula. I did the same and ended up getting this.

$$\tan(\theta)=-\cot(\theta-90^\circ)$$

which is wrong according to the book.

It says the answer should be…

$$\tan(\theta-90^\circ)=-\cot(\theta)$$

I do not where am I going wrong. What is incorrect in my approach? What is the flaw? Please help me track it down.

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marked as duplicate by symplectomorphic, Anurag A, Guy Fsone, José Carlos Santos, Especially Lime Nov 15 '17 at 9:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ There is no flaw. Book and you have the same answer. Just use $\tan x=\frac{1}{\cot x}$ in what you have arrived at. $\endgroup$ – Anurag A Nov 15 '17 at 8:14
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    $\begingroup$ You already asked this question. You should wait for an answer to the other question. $\endgroup$ – symplectomorphic Nov 15 '17 at 8:15
  • $\begingroup$ @symplectomorphic I am sorry. I couldn't wait for the answer so I asked it again. $\endgroup$ – Saksham Sharma Nov 15 '17 at 8:16
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    $\begingroup$ @AnuragA: post that comment as an answer on the other question. $\endgroup$ – symplectomorphic Nov 15 '17 at 8:18
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Use the reciprocal identity... $$\tan(x)=\frac{1}{\cot(x)}$$

and substitute it in the respective variables to arrive at the desired answer.

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