In an equilateral spherical triangle, show that $\sec A=1+\sec a$ Q. In an equilateral spherical triangle, show that $\sec A=1+\sec a$
So $A$ is the vertex or the angle of the triangle and $a$ is the side of the equilateral spherical triangle.
I started off the proof by using the law of cosines:
$$\cos(a)-\cos(a)\cos(a)=\sin(a)\sin(a)\cos(A)$$
and after simplifying it a bit, I obtained:
$$\cos(a)-\cos^2(a)=\sin^2(a)\cos(A)$$
I replaced $\sin^2(a)$ with $(1-\cos^2(a))$.
Then I obtained:
$$\cos(a)-\cos^2(a)=(1-\cos^2(a))\cos(A)$$
and I realized on the left side, I can pull a $\cos(a)$. So through factoring:
$$\cos(a)(1-\cos(a))=(1-\cos^2(a))\cos(A)$$
Either I'm not seeing it but I do not how to proceed after this. If anyone can help, I'd like that. Thanks!

Let $\,x:=\cos(a)\,$ and $\,X:=\cos(A).\,$ Then
$$ 1+\sec(a)-\sec(A) = 1+\frac1x-\frac1X = \frac{-x+X+xX}{xX}.$$
$$ \cos(a)(1-\cos(a))-(1-\cos^2(a))\cos(A)=x(1-x)-(1-x^2)X=(x-1)(-x+X+xX).$$
Thus, if $\,x\ne 1,\,$ then
$\, \cos(a)(1-\cos(a))=(1-\cos^2(a))\cos(A)\, $ implies that
$\, 1+\sec(a)=\sec(A).$
