In an equilateral spherical triangle, show that $\sec A=1+\sec a$

Question

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!

Answer 1

lambda, mu are lattitude and longitude. $R = 1$.

$O = (0,0,0)$

$A = (\cos \lambda_a \cos \mu_a, \cos \lambda_a \sin \mu_a, \sin \lambda_a)$

$B = (\cos \lambda_b \cos \mu_b, \cos \lambda_b \sin \mu_b, \sin \lambda_b)$

$C = (\cos \lambda_c \cos \mu_c, \cos \lambda_c \sin \mu_c, \sin \lambda_c)$

The planes $AOB$ and $AOC$ have the line $OA$ in common. Let $\alpha = \angle (AOB, AOC)$.

$OB \cdot OC = \cos \beta$.

You claim that $\frac{1}{\cos \alpha} = 1 + \frac{1}{\cos \beta}$ (*)

Angle between normals implies vectorial product: $\alpha = \angle (OA \times OB, OA \times OC)$.

$\cos \alpha = (OA \times OB) \cdot (OA \times OC)$

$(*) \Rightarrow \cos \beta = \cos \alpha \cos \beta + \cos \alpha \Rightarrow \cos \beta (1 - \cos \alpha) = \cos \alpha$

$\cos \alpha = b_x c_x + b_y c_y + b_z c_z$

$A \times B = (a_y b_z - a_z b_y, - a_z b_x + a_x b_z, a_x b_y - a_y b_x)$

$A \times C = (a_y c_z - a_z c_y, - a_z c_x + a_x c_z, a_x c_y - a_y c_x)$

$\cos \beta = (a_y b_z - a_z b_y)(a_y c_z - a_z c_y) + (- a_z b_x + a_x b_z)(- a_z c_x + a_x c_z) + (a_x b_y - a_y b_x)(a_x c_y - a_y c_x)$

$\cos \beta = (a_y^2 b_z c_z - a_y a_z b_z c_y - a_y a_z b_y c_z + a_z^2 b_y c_y) + (a_z^2 b_x c_x - a_x a_z b_x c_z - a_x a_z b_z c_x + a_x^2 b_z c_z) + (a_x^2 b_y c_y - a_x a_y b_y c_x - a_x a_y b_x c_y + a_y^2 b_x c_x)$

$\cos \beta = a_y^2 (\cos \alpha - b_y c_y) + a_z^2 (\cos \alpha - b_z c_z) + a_x^2 (\cos \alpha - b_x c_x) + (- a_y a_z b_z c_y - a_y a_z b_y c_z) + (- a_x a_z b_x c_z - a_x a_z b_z c_x) + (- a_x a_y b_y c_x - a_x a_y b_x c_y)$

$\cos \beta = \cos \alpha - a_x^2 b_x c_x - a_y^2 b_y c_y - a_z^2 b_z c_z + (- \cos \lambda_a \sin \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \sin \mu_c - \cos \lambda_a \sin \mu_a \sin \lambda_a \cos \lambda_b \sin \mu_b \sin \lambda_c) + (- \cos \lambda_a \cos \mu_a \sin \lambda_a \cos \lambda_b \cos \mu_b \sin \lambda_c - \cos \lambda_a \cos \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \cos \mu_c) + (- \cos \lambda_a \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \sin \mu_b \cos \lambda_c \cos \mu_c - \cos \lambda_a \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \cos \mu_b \cos \lambda_c \sin \mu_c)$

$\cos \beta = \cos \alpha - a_x^2 b_x c_x - a_y^2 b_y c_y - a_z^2 b_z c_z + \cos \lambda_a [(- \sin \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \sin \mu_c - \sin \mu_a \sin \lambda_a \cos \lambda_b \sin \mu_b \sin \lambda_c) + (- \cos \mu_a \sin \lambda_a \cos \lambda_b \cos \mu_b \sin \lambda_c - \cos \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \cos \mu_c) + (- \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \sin \mu_b \cos \lambda_c \cos \mu_c - \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \cos \mu_b \cos \lambda_c \sin \mu_c)]$

I may have not defined right.

Answer 2

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 $\, 1+\sec(a)=\sec(A).$