1. ## inverse trigonometric function

The problem is

Show that Sin-1(12/13) + Cos-1(4/5) + Tan-1(63/16) = Pi

I converted Sin-1(12/13) as Tan-1(12/5)

Cos-1(4/5) as Tan-1(3/4)

Then I took the above problem as

Tan-1(12/5) + Tan-1(3/4) = Tan-1{[12/5 + 3/4]/[1-(12/5)(3/4)]} = Tan-1(-63/16) = - Tan-1(63/16)

The above problem becomes zero (0).

How I could conclude that the answer is Pi?

or What is wrong with my working.

Kindly guide me.

with warm regards,

Aranga

2. ## Re: inverse trigonometric function

$\displaystyle \tan ^{-1}a+\tan ^{-1}b\overset{\text{??}}{=}\tan ^{-1}\left(\frac{a+b}{1-a b}\right)$

3. ## Re: inverse trigonometric function

Thanks. Now I revisited the formula given in the book.

They have mentioned that

tan-1x + tan-1y = tan-1{[x + y]/[1-xy]} only when xy<1.

Why such condition they have given I did not understand.

in this case, xy is greater than 1. So it fails. That much I understand.

Now I do not know how to solve this problem or how to proceed further.

Kindly guide me.

with warm regards,

Aranga

4. ## Re: inverse trigonometric function

if $\displaystyle x y>1$

$\displaystyle \tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) + \pi$

P.S. I am not sure if this is correct

5. ## Re: inverse trigonometric function

Thanks. As you said If I put I am getting the answer.

Whether it is correct?

If it is so, how it is correct?

with warm regards,

Aranga

6. ## Re: inverse trigonometric function

I think I would arrange the equation as:

$\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)$

Take the sine of both sides (apply the angle sum identity for sine on the left and the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ on the right):

$\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)$

$\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}$

$\displaystyle \frac{48+15}{65}=\frac{63}{65}$

$\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark$

7. ## Re: inverse trigonometric function

please see the post of IDEA.

Whether what was said is correct?

8. ## Re: inverse trigonometric function

Originally Posted by arangu1508
please see the post of IDEA.

Whether what was said is correct?
It looks correct to me, and I think we only require $\displaystyle xy\ne1$.

9. ## Re: inverse trigonometric function

thanks.

It was very useful.

10. ## Re: inverse trigonometric function

Originally Posted by MarkFL
I think I would arrange the equation as:

$\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)$

Take the sine of both sides (apply the angle sum identity for sine on the left and the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ on the right):

$\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)$

$\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}$

$\displaystyle \frac{48+15}{65}=\frac{63}{65}$

$\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark$
This argument is not valid.

To see this, start with the following equation which is obviously not correct

$\displaystyle \sin ^{-1}\frac{12}{13}+\cos ^{-1}\frac{4}{5}=\tan ^{-1}\frac{63}{16}$

and apply $sin$ to both sides to get

$\displaystyle \frac{63}{65}=\frac{63}{65}$

so taking sin of both sides and getting equal answers does not prove the original assertion

11. ## Re: inverse trigonometric function

Three angles in $(0,\pi)$ add up to $\pi$ implies there exists a triangle with those angles as measures. Can that fact be used to prove the assertion?

12. ## Re: inverse trigonometric function

Yes...I bungled all over the place in this thread.

13. ## Re: inverse trigonometric function

if x > 0 , y > 0 , and xy > 1 then

$\displaystyle \tan ^1x+\tan ^{-1}y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)+\pi$

To prove this let $\displaystyle a=\tan ^{-1}x$, $\displaystyle b=\tan ^{-1}y$ where $\displaystyle 0<a,b<\pi /2$

$\displaystyle \tan (a+b)=\frac{x+y}{1-x y}<0$

therefore $\displaystyle \pi /2 < a+b < \pi$

this gives

$\displaystyle \tan ^{-1 }(\tan (a+b))=a+b-\pi$

done.

Another proof, using calculus of two variables, let

$\displaystyle g(x,y)=\tan ^1x+\tan ^{-1}y-\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ , $\displaystyle x y\neq 1$

verify that both partial derivatives are zero in each of the three regions where $\displaystyle g$ is defined.

therefore $\displaystyle g$ is constant. Check one point in each region to find out that the constants are $\displaystyle 0,\pi ,- \pi$

14. ## Re: inverse trigonometric function

Thank you very much. It is very useful. I am able to follow the first proof.

So, In my case, x > 0, y > 0 and xy > 1,

I can straight away conclude that

tan-1x + tan-1y = tan-1{[x + y]/[1-xy]} + pi

with warm regards,

Aranga