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\begin{align} \tan(A) + \cot(A) & = 2 \text{cosec}(2A)\\ \tan(45^{\circ}+A^{\circ}) - \tan(45^{\circ}-A^{\circ}) & = 2 \tan(2A^{\circ})\\ \text{cosec}(2A) + \cot(2A) & = \cot(A) \end{align}

I have got all the formulas that I need but I just couldn't solve these. Some help, please?

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For the second one: $\tan(u+v)=\frac{\tan\,u+\tan\,v}{1-\tan\,u\tan\,v}$ and $\tan 45^\circ=1$. – J. M. Aug 7 '11 at 13:43
@Sophia: Do you mean $\csc(2\textrm A)$ for cosec($2\textrm A$)? – night owl Aug 8 '11 at 10:00
I have solved the 1st and 3rd question, but I still can't get the second question! Anymore further hints please? – Sophia Aug 8 '11 at 13:14
I gave a formula you can use for both the left and right hand sides... $u+u=2u$, right? – J. M. Aug 9 '11 at 8:02

1) $\tan a+\cot a=\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a}$. Bring to a common denominator and use the identities you know. (remember that $\csc x=\frac{1}{\sin x}$).
2) Use the same hint from 1. Use the formulas for $\sin(a+b)$ and $\cos(a+b)$.
3) Same hint from 1.

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Thanks for the hints. But I still got stuck at the part where I reached ( 1/cosAsinA) for the first question. How do I go on from there? – Sophia Aug 7 '11 at 13:37
@Sophia: $\sin\,2u=2\sin\,u\cos\,u$ – J. M. Aug 7 '11 at 13:41
@J. M. Oh! I get it now, must have a glitch in my brain for not seeing the obvious, thanks a lot! – Sophia Aug 7 '11 at 13:44

EDITED. In general there are two different techniques we can use to prove a trigonometric identity $A=B$. One is to transform one side into the other: $$A=A_1=A_2=\dots =A_n=B.$$ The other is to look at the identity $A=B$ as a whole and convert it into an equivalent one and repeat the process until one known identity is found: $$A=B\Leftrightarrow A'=B'\Leftrightarrow A''=B''\Leftrightarrow\dots\Leftrightarrow A^{*}=B^{*}.$$ The following hints are intended for proving your $3^{rd}$ identity by this second technique. Use $$\frac{\cos 2A}{\sin 2A}=\frac{2\cos ^{2}A-1}{2\sin A\cos A}=\frac{% \cos A}{\sin A}-\frac{1}{2\sin A\cos A}$$ to obtain $$\csc 2A+\cot 2A=\cot A\Leftrightarrow \frac{1}{2\cos A}+\cos A-\frac{1}{% 2\cos A}=\cos A.$$

Added. Proof:

$$\csc 2A+\cot 2A=\cot A\tag{1}$$

$$\begin{eqnarray*} &\Leftrightarrow &\frac{1}{\sin 2A}+\frac{\cos 2A}{\sin 2A}=\frac{\cos A}{% \sin A} \\ &\Leftrightarrow &\frac{\sin A}{\sin 2A}+\frac{\cos 2A}{\sin 2A}\sin A=\cos A \\ &\Leftrightarrow &\frac{1}{2\cos A}+\left( \frac{\cos A}{\sin A}-\frac{1}{% 2\sin A\cos A}\right) \sin A=\cos A \\ &\Leftrightarrow &\frac{1}{2\cos A}+\left( \cos A-\frac{1}{2\cos A}\right) =\cos A \\ \end{eqnarray*}$$

$$\Leftrightarrow \cos A=\cos A\tag{2},$$

which is an identity. Thus $(1)$ is also an identity.

Your $1^{st}$ identity can be proved by the first technique:

$$\begin{eqnarray*} \tan A+\cot A &=&\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin ^{2}A+\cos ^{2}A}{\cos A\sin A} \\ &=&\frac{1}{\cos A\sin A}=\dfrac{1}{\dfrac{\sin (2A)}{2}}=\cdots \end{eqnarray*}$$

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For the second one:

  • $2)$ Call $\alpha = \frac{\pi}{4} +A$ and $\beta= \frac{\pi}{4}-A$. So you have $\alpha + \beta = \frac{\pi}{2} \Rightarrow \cot(\alpha+\beta) = 0$. From this you have $$\frac{1}{\tan(\alpha+\beta)} = \frac{1 - \tan{\alpha}\tan{\beta}}{\tan{\alpha} + \tan{\beta}}=0 \Rightarrow \tan{\alpha} \cdot \tan{\beta}=1$$ Now note that $\alpha - \beta= 2A$ and so we again have \begin{align*} \tan(\alpha-\beta) &= \frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\cdot \tan(\beta)} \\ &= \frac{\tan(\alpha)-\tan(\beta)}{2} \\ \Rightarrow \ 2\tan(2A)&= \tan(\alpha)- \tan(\beta) \end{align*}

  • $3)$ You have : \begin{align*} \csc{2A} + \cot{2A} &= \frac{1}{\sin{2A}} + \frac{\cos{2A}}{\sin{2A}} \\ & = \frac{1+\cos{2A}}{\sin{2A}} \\ &= \frac{1+\cos^{2}{A} - \sin^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} \\ &= \frac{2 \cdot\cos^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} = \cot(A)\end{align*}

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