Eliminate ${\theta, \varphi}$ from $\tan\theta+\tan\varphi=a$, $\sec\theta+\sec\varphi=b$, $\csc\theta+\csc\varphi=c$ Here is the problem:
Eliminate $\theta, \varphi$ from the equations
$$\tan\theta+\tan\varphi=a$$
$$\sec\theta+\sec\varphi=b$$
$$\csc\theta+\csc\varphi=c$$
What I have so far.
$$\tan \theta=a-\tan\varphi$$
$$\sec \theta=b-\sec\varphi$$
$$\csc \theta=c-\csc\varphi$$
And using
$$\tan \theta \csc\theta =\sec\theta$$ gives us
$$c\tan\varphi+a\csc\varphi-2\sec \varphi=ac-b$$
And also
$$\tan^2\theta+1=\sec^2\theta$$
gives
$$2a\tan\varphi-2b\sec\varphi=a^2-b^2$$
Thus we have two linear relations. But I cannot think of any more properties to use. Things like $\sin^2+\cos^2=1$ give a big mess. So what are the end steps ?
Update: The method used in this question: How to eliminate $\theta$ & $\phi$ from above equations
will work and solves the problem.
Following that method gives me
$$[(b^2-a^2)^2+4a^2]a^2c=4a^2[c(a^2+b^2)-2ab]$$
What a fantastic relation !

Disclosure: This almost seems to be the answer that @Rene Schipperus found (originally by @John Bentin on this post), but has some variations. So I am posting it. Although some lines are exactly as the previous answer, I will include them here for the sake of completeness.

$$\tan\theta+\tan\varphi=a$$
$$\sec\theta+\sec\varphi=b$$
$$\csc\theta+\csc\varphi=c$$
Changing the shape of these equations, $$\sin\theta\cos\varphi+\sin\varphi\cos\theta=a\cos\theta\cos\varphi\tag1$$
$$\cos\theta+\cos\varphi=b\cos\theta\cos\varphi\tag2$$
$$\sin\theta+\sin\varphi=c\sin\theta\sin\varphi\tag3$$
Further moulding,
$$2\sin\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta+\varphi}2\right)=a\cos\theta\cos\varphi\tag4$$
$$2\cos\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)=b\cos\theta\cos\varphi\tag5$$
$$2\sin\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)=c\sin\theta\sin\varphi\tag6$$
Dividing (4) by (5),
$$\frac{\sin\left(\frac{\theta+\varphi}2\right)}{\cos\left(\frac{\theta-\varphi}2\right)}=\frac ab\tag7$$
Multiply (6) by (7) and 1/(7) respectively,
$$2\sin^2\left(\frac{\theta+\varphi}2\right)=\frac{ac}b\sin\theta\sin\varphi$$
$$2\cos^2\left(\frac{\theta-\varphi}2\right)=\frac{bc}a\sin\theta\sin\varphi$$
From the last two we get,
$$1-\cos(\theta+\varphi)=\frac{ac}b\sin\theta\sin\varphi\tag8$$
$$1+\cos(\theta-\varphi)=\frac{bc}a\sin\theta\sin\varphi\tag9$$
(9) - (8)
$$\cos(\theta+\varphi)+\cos(\theta-\varphi)=\left(\frac{bc}a-\frac{ac}b\right)\sin\theta\sin\varphi$$
$$2\cos\theta\cos\varphi=\frac{c(b^2-a^2)}{ab}\sin\theta\sin\varphi$$ $$\tan\theta\tan\varphi=\frac{2ab}{c(b^2-a^2)}=\lambda\quad\text{  (say)}$$
Ok, leave it here and let's move back to first few equations.
Dividing (6) by (5),
$$\tan\left(\frac{\theta+\varphi}2\right)=\frac cb\tan\theta\tan\varphi=\frac cb\lambda$$
Now using, $$\tan(\theta+\varphi)=\frac{2\tan\left(\frac{\theta+\varphi}2\right)}{1-\tan^2\left(\frac{\theta+\varphi}2\right)}$$
$$\frac{\tan\theta+\tan\varphi}{1-\tan\theta\tan\varphi}=\frac{2\tan\left(\frac{\theta+\varphi}2\right)}{1-\tan^2\left(\frac{\theta+\varphi}2\right)}$$
$$\frac a{1-\lambda}=\frac{2\dfrac cb\lambda}{1-\left(\dfrac cb\lambda\right)^2}$$
Further simplifications lead to, $$c[(b^2-a^2)^2-4a^2]=4[c(b^2-a^2)-2ab]$$ and finally, $$(b^2-a^2)^2c=4b^2c-8ab$$ which is indeed the answer obtained using the other method.
An alternate form is $$\boldsymbol{(2b-a^2+b^2)(2b+a^2-b^2)c=8ab}$$ which seems better!
