# Proving basic/standard trigonometric identities

How to prove the following trigonometric identities ?

1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$

I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?

2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$

For this one I tried substituting them with the sides of the triangle, but not successful to the final result.

These are not my homework, I am trying to learn maths almost on my own, so ...

• @Tretwick Marian: The fist one should be $\tan(\alpha) \tan(\beta) = 1$, then $\alpha + \beta = \frac{\pi}{2}$. The second one should be $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$. – user17762 Nov 27 '10 at 8:08
• Unless there are restrictions on $\alpha$ and $\beta$, the first conclusion is not correct; it should be $\alpha+\beta=\pi/2+n\pi$ for some integer $n$. – Hans Lundmark Nov 27 '10 at 14:09
• The first equality is not an identity. Perhaps the title should be changed. – Américo Tavares Nov 27 '10 at 16:00

$$\tan \alpha \tan \beta =1\Leftrightarrow \dfrac{\sin \alpha }{\cos \alpha }% \dfrac{\sin \beta }{\cos \beta }=1$$

Multiplying by $\cos\alpha\cos\beta\ne 0$, gives

$$\sin \alpha \sin \beta -\cos \alpha\cos \beta =0 \Leftrightarrow \cos (\alpha +\beta )=0$$

This is equivalent to

$$\alpha +\beta =\dfrac{\pi }{2}+n\pi,\qquad (\ast)$$

to which we still have to add the condition written above ($\cos\alpha\cos\beta\ne 0$), which means the constraint

$$\alpha,\beta\ne\dfrac{\pi}{2}+n\pi,\qquad (\ast\ast)$$

where $n$ is an integer.

Note: In the original equation $\tan \alpha \tan \beta =1$, neither $\alpha$ nor $\beta$ can be zero. The combined conditions $(\ast)$ and $(\ast\ast)$ assures that.

The identity $$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\qquad \theta \neq (2n+1)\dfrac{\pi}{2}$$

is equivalent to $$\sin ^{2}\theta +\cos ^{2}\theta =1.$$

Indeed, if

$$\theta \neq (2n+1)\dfrac{\pi }{2}\Leftrightarrow \sin \theta \neq \pm 1 \Leftrightarrow \dfrac{\pm1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }\neq 0\Leftrightarrow \pm\sec \theta -\tan \theta \neq 0,$$

then

$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta -\tan \theta \right) =1$$

$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$

• There are several errors in the first computation; for example, in the second step you have three $\alpha$'s and one $\beta$. – Hans Lundmark Nov 27 '10 at 14:05
• @Hans Lundmark, Many thanks! I have corrected them. – Américo Tavares Nov 27 '10 at 14:47
• Fine, except that there is one step where equivalence doesn't hold (see the comments to user3971's answer), and you've forgotten to add $n\pi$ at the end. – Hans Lundmark Nov 27 '10 at 15:01
• @Hans Lundmark, I corrected once more the first computation. – Américo Tavares Nov 27 '10 at 15:35
• @Tretwick Marian, concerning the first equation, please look at the new solution I wrote after you has accepted my answer. – Américo Tavares Nov 27 '10 at 16:39

When I need to prove trigonometric identities, I tend to use complex exponentials. For instance, to prove your first identity observe that \begin{align} \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} \quad \text{and} \quad \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \end{align} so \begin{align} \tan \theta = \frac{1}{i} \frac{e^{i\theta} - e^{-i\theta} }{e^{i \theta} + e^{-i \theta}}. \end{align} We calculate \begin{align} 1 = \left( \frac{1}{i} \frac{e^{i\alpha} - e^{-i\alpha} }{e^{i \alpha} + e^{-i \alpha}} \right) \left(\frac{1}{i} \frac{e^{i\beta} - e^{-i\beta} }{e^{i \beta} + e^{-i \beta}} \right) = - \frac{(e^{2 i \alpha} - 1)(e^{2 i \beta} - 1)}{(e^{2 i \alpha} + 1)(e^{2 i \beta} + 1)} \end{align} Hence, \begin{align} (e^{2 i \alpha} + 1)(e^{2 i \beta} + 1) = - (e^{2 i \alpha} - 1)(e^{2 i \beta} - 1). \end{align} or $e^{2 i (\alpha + \beta)} + 1 = 0$, which implies that $\alpha + \beta = \frac{\pi}{2}$ by Euler's equation $e^{\pi i} + 1 = 0$, provided that we consider only angles in the fundamental region $[-\frac{\pi}{2}, \frac{\pi}{2}]$. A similar calculation works for your second identity.

1) tan(a)tan(b) = 1 is equivilant to sin(a)sin(b) = Cos(a)cos(b) **using the fact that tan is sine divided by cosine. Next use cos(a+b) = cos(a)cos(b) -sin(a)sin(b)

which implies sin(a)sin(b) = cos(a)cos(b) -cos(a+b)

Use this on the left hand side of ** to get

                              cos(a+b) = 0


which implies a+b = Pi/2 or -Pi/2

• -1: First, tan(a)tan(b) = 1 and sin(a)sin(b) = cos(a)cos(b) are not equivalent. Second cos(a+b)= 0 does not imply a+b = pi/2 or -pi/2. – Aryabhata Nov 27 '10 at 8:33
• btw, you have the right idea :-). So please edit the answer to correct the mistakes so that I can upvote. – Aryabhata Nov 27 '10 at 8:41
• @Moron: I am not getting your first comment.Please explain :) – Quixotic Nov 27 '10 at 10:32
• @Debanjan: The implication only goes one way; for example, if $a=0$ and $b=\pi/2$ the equation with sin & cos is satisfied, but not the one with tan. And $\cos x=0$ iff $x=\pi/2+n\pi$ where $n$ is an arbitrary integer (not just $n=0$ or $n=-1$). – Hans Lundmark Nov 27 '10 at 14:58
• Hmm,...,got it. – Quixotic Nov 27 '10 at 18:57

For (2)

$$\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}= \frac{1 + \sin \theta}{\cos \theta}$$ $$= \frac{(1 + \sin \theta)\cdot (1 - \sin \theta)}{\cos \theta \cdot(1 - \sin \theta)}\text{ [Multiplying both sides by } (1 - \sin \theta)\text{]}$$ $$= \frac{\cos^2 \theta}{ \cos \theta - \sin \theta \cdot \cos \theta} = \frac{1}{\sec \theta - \tan \theta} \text{ (Q.E.D) }$$