# Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum)

I'm looking for a specific clarification for a part of the solution in proving the following identity. $$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$$

Here I'm taking, $$\theta=\tan^{-1}m;$$ $$-\pi/2<\theta<\pi/2$$

so I get $$\tan\theta=m$$---(1)

I need to find $$\sin \theta$$and $$\cos \theta$$ in terms of m

By Trigonometric identity I can easily derive, $$\cos \theta$$

$$\tan^2\theta+1=\sec^2\theta$$

$$\cos^2\theta=\frac{1}{m^2+1}$$

$$\cos\theta=+\sqrt\frac{1}{m^2+1}$$ ( here only plus due to range of $$\theta$$)

Now if I deduce $$\sin\theta$$ from,

$$\sin^2\theta+\cos^2\theta=1$$

I get, $$\sin\theta=\pm\sqrt\frac{m^2}{m^2+1}$$ ( I have to take $$\pm$$ because of range of $$\theta$$)

But if I deduce $$\sin\theta$$ from (1)

I get, $$\sin\theta=\frac{m}{\sqrt {m^2+1}}$$

Which of the following method is correct to find $$\sin\theta$$? Please help me. Thank you!

P.S. I'm not interested in the solution. What I need to know is how to find $$\sin\theta$$

• How did you get the positive sign for $\sin \theta$ using 1)? – Kavi Rama Murthy Mar 20 at 8:55
• since equation (1) means $\tan\theta=\frac{\sin\theta}{\cos\theta}$, I substituted the $cos\theta$value I obtained there. – emil Mar 20 at 10:00
• You need appropriate restrictions on $m$ and $n$. For example, the formula is false in the case $m=-\surd3$ with $n=1/\surd3$. – John Bentin Mar 20 at 11:36

Using this

We need $$\tan^{-1}m+\tan^{-1}n\ge0$$ to admit the equality

$$\tan^{-1}m+\tan^{-1}n\ge0\iff\tan^{-1}m>-\tan^{-1}n=\tan^{-1}(-n)$$ $$\iff m\ge-n\iff m+n\ge0$$

Then if $$\tan^{-1}x=y,-\dfrac\pi20$$

$$x=\tan y\implies\cos y=\dfrac1{\sqrt{1+x^2}}$$

$$\tan^{-1}x=\text{sign of}(x)\cdot\cos^{-1}\dfrac1{\sqrt{1+x^2}}$$

I don't have means to put up the image for my plot so bear with me.

Take a right-angled triangle $$ABC$$ such that AB is the height $$=m$$ and BC is the base $$=1$$. $$\widehat{ABC}=90^\circ$$ and $$\widehat{ACB}=\tan^{-1} m$$

Let $$AC=y$$. Draw a line $$AE$$ perpendicular to $$AC$$ at $$A$$ equal to $$ny$$. $$\widehat{EAC}=90^\circ$$ and $$\widehat{ACE}=\tan^{-1} \dfrac{ny}{y}=\tan^{-1} n$$. Additionally, let $$EC=z$$

Extend $$AB$$ past $$A$$. Draw a perpendicular from E to that line and name that intersection $$D$$. $$\widehat{EAD}=\widehat{ACB}$$. This is because of the right angle $$\widehat{EAC}$$. Hence $$\triangle ABC \sim \triangle EDA$$. This equality follows:

$$\dfrac{AB}{AC}=\dfrac{ED}{EA}$$

$$\dfrac{m}{y}=\dfrac{ED}{ny}$$

$$ED=mn$$

Draw a perpendicular to $$BC$$ at $$C$$ and extend $$DE$$ past $$E$$. Label the point where these lines meet as $$F$$. We have created a rectangle in $$DBCF$$, thus $$DF=BC=1$$:

$$EF+DE=1$$

$$EF=1-mn$$

Additionally, $$DF \parallel BC$$ hence $$\widehat{ECB}=\widehat{CEF} \text{ Alternate angles}$$

$$\widehat{ECB}=\widehat{ACB}+\widehat{ECA}=\tan^{-1} m+\tan^{-1} n$$

$$\widehat{CEF}=\cos^{-1} \dfrac{EF}{EC}$$

$$\dfrac{EF}{EC}=\dfrac{1-mn}{z}$$

Remember now all the right-angled triangles we drew.

$$z^2=(ny)^2+y^2=y^2(n^2+1)$$

$$y^2=m^2+1$$

$$z^2=(n^2+1)(m^2+1)$$

$$z=\sqrt{(n^2+1)(m^2+1)}$$

$$\dfrac{EF}{EC}=\dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$$

$$\widehat{CEF}=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$$

But $$\widehat{ECB}=\widehat{CEF}$$, hence:

$$\boxed{\tan^{-1} m+\tan^{-1} n=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}}$$

Edit: the formula kinda breaks down when $$m$$ and $$n$$ don't have the same sign i.e $$n \lt 0 \lt m$$ for example. But they won't be that different. For example, using $$-\sqrt{3}$$ and $$\frac{1}{\sqrt{3}}$$ yields $$-30^\circ$$ and $$30^\circ$$ on the left and right respectively. But they do have the same cosine value meaning they're not too distinct.

• what if m is negative will your statement be correct length AB=m?( second statement of your proof) – emil Mar 20 at 10:04
• Yes. Just flipped over. I think my diagram would extend for positive and negative acute angles, from which all other quadrants can follow. – Nεo Pλατo Mar 20 at 10:20
• And besides, if you check Wikipedia their proof of the sine addition formula is geometric, from which all other angles can follow. Also, if $m\ge \frac{1}{n}$, then $\tan^{-1}m + \tan^{-1} n \ge 90^\circ$ and $1 \le mn$, $1-mn \le 0$ which does correspond to the cosine of an obtuse angle. – Nεo Pλατo Mar 20 at 10:51
• Sorry as you pointed out the answer is what you have written.. I just now corrected it. But how can you address the sine issue? – emil Mar 20 at 10:51
• It's completely correct. Both of them. It's just that the second one came immediately from a square root. That can be a problem sometimes. The best thing for that is to decide for yourself which quadrant you're working with and move on with it. Like the way I know mine is obviously in acute angles. But the determination is important. For example if you used that Pythagorean identity with a negative cosine to work out sine then the positive value is for the 2nd quadrant angle and the negative the 3rd and so on. – Nεo Pλατo Mar 20 at 10:56