Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ...

share|improve this question
@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
@Tretwick Marian: If this is homework, kindly tag it as homework so that we can help you. What have you tried so far? –  user17762 Nov 27 '10 at 8:08
You've asked a lot of questions on this site. You might want to consider registering. –  Guess who it is. Nov 27 '10 at 8:24
J.M: I have,care checking my profile ? ;) –  VelvetThunder Nov 27 '10 at 8:26
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

4 Answers 4

up vote 5 down vote accepted

Added (and corrected twice):

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

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

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


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

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

share|improve this answer
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.

share|improve this answer

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

share|improve this answer
-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 :) –  VelvetThunder 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. –  VelvetThunder Nov 27 '10 at 18:57

For (2)

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.