# How to eliminate $\theta$ & $\phi$ from above equations

Eliminate $\theta$ & $\phi$ from above equations.

$\cos(\theta)+\cos(\phi)=a$--------------(say Eq1)

$\cot(\theta)+\cot(\phi)=b$--------------(say Eq2)

$\operatorname{cosec}(\theta)+\operatorname{cosec}(\phi)=c$--------------(say Eq3)

What I tried:

From: $(Eq3)^2 -(Eq2)^2$

$c^2-b^2-2=2(\operatorname{cosec}(\theta)\cdot\operatorname{cosec}(\phi)-\cot(\theta)\cdot\cot(\phi))$

From: $Eq1\times Eq3$

$ca-b=\frac{\sin(\theta)\cos(\theta)+\sin(\phi)\cos(\phi)}{sin(\theta)\cdot\sin(\phi)}$

From these two derived expressions I was able to simplify it further.( But could not completely eliminate). I'm not sure whether my approach is correct. Can you please give me a hint. Thanks.

• You have three equations and two variables to eliminate? – Dr. Sonnhard Graubner Jul 3 '18 at 9:55
• yes, It is required to form a single equation that doesn't have $\phi$ & $\theta$ – emil Jul 3 '18 at 10:52

We may write these equations as$$\cos \theta+\cos\phi=a,\qquad$$$$\cos\theta\sin\phi+\sin\theta\cos\phi=b\sin\theta\sin\phi,\qquad$$$$\qquad\sin\theta+\sin\phi=c\sin\theta\sin\phi.$$Let us write $\alpha:=\frac12(\theta+\phi)$ and $\beta=\frac12(\theta-\phi)$. Then the above equations may be written, using standard trigonometric formulae, as$$2\cos\alpha\cos\beta=a,\qquad\qquad\qquad\qquad(1)$$$$4\sin\alpha\cos\alpha=b(\cos2\beta-\cos2\alpha),\qquad$$$$4\sin\alpha\cos\beta=c(\cos2\beta-\cos2\alpha).\qquad$$Consider first the special case when $\cos2\beta=\cos2\alpha.$ This is when either $\theta$ or $\phi$ is a multiple of $\pi$. But then the original equations would be undefined; so we can rule this case out. Thus we may divide the last two displayed equations to obtain$$c\cos\alpha=b\cos\beta.\qquad\qquad\qquad\qquad(2)$$By subtracting the same equations, writing $\cos2\beta-\cos2\alpha=2(\cos\beta-\cos\alpha)(\cos\beta+\cos\alpha)$, and dividing through by $\cos\beta-\cos\alpha$ (which must be nonzero in the present case), we get$$2\sin\alpha=(c-b)(\cos\alpha+\cos\beta).\qquad(3)$$ Now substituting for $\cos\beta$ from eqn $2$ into eqns $1$ and $3$, using the fact that $\cos^2\alpha+\sin^2\alpha=1$, and some straightforward algebra yields$$a(c^2-b^2)^2=4b(2c-ab).$$
$$\cos(x)=\frac{1-\tan(x/2)^2}{1+\tan(x/2)^2}$$ and $$\cot(x)=\frac{1}{\tan(x)}$$ and $$\tan(x)=\frac{2\tan(x/2)}{1-\tan^2(x/2)}$$