Why is the polar triangle useful in spherical geometry? We can solve many problems in spherical geometry by using the polar triangle. I am looking for an intuition why (and when) this is easier than working in the original triangle.
 A: I think that your question is interesting. Polar triangles make easier some trigonometrical derivations as indicated by 
the Wikipedia page . A good reference for this topic is the very
old, but very good book Spherical Trigonometry by Todhunder.
This is a long answer and I will do in two parts:


*

*Geometrical Insights: The concept of duality

*Applications:  


*

*Linear Algebra.

*Tensor Analysis.

*Functional Analysis.



Geometrical Insight: The concept of duality.
Besides the fact that polar triangles make some trigonometrical derivations easier they present a good example for the theory of dualitites.
There is a discussion on the concept of duality here .
It is clear that to each point of the sphere there correspond an equator. That point can be called a north pole. Likewise to each equator there corresponds a north pole. So there is a duality between points and their equators (points and lines in the sphere). When we consider a set of points and its dual (their equators) we can jump to more general examples. An interesting example is that of a segment and its dual a lune. When the segment is swept a lune is created. For example see this figure:

A segment and its dual a lune. Think of the firsts point of the
  segment as the point $(0,1,0)$ in the $Y$ axis with green color. The equator
corresponding to this point is the intersection of the $XZ$ plane with the
sphere, a green circle.  Now if the segment is a small piece of arc rotating
30 degrees from the axis $Y$ toward the axis $X$, as we move along this
path, the vertical circle in the $XZ$ plane starts rotating with the
segment at the same speed toward a blue color. The shaded area on the
sphere is a lune
We want to go further and ask what is the dual of a triangle in a sphere?
First look at this figure:

It seems that the blue triangle is dual of the red. It is actually  a dual but this particular duality is not very interesting. The antipode triangle is like a spherical mirror image of the prime triangle and it is congruent to the prime but inverted.
A triangle which is more interesting is the polar triangle. Here is a figure and how it is constructed:

Given an spherical triangle (red) with angles $\alpha$, $\beta$, and
    $\gamma$, and sides $a$, $b$, and $c$, the corresponding polar triangle
    is found (blue) with angles $\alpha'$, $\beta'$, and $\gamma'$, sides
  $a',b'$ and $c'$. The sides of the polar triangle join the vertices along
great circles. The vertex at $\alpha' $ is the north pole of the equator
for the side $a$, the vertex at $\beta'$ is the north pole for the equator having
the side $b$, and the vertex at $\gamma'$ is the north pole for the equator
having the side $c$.
What is interesting is that the polar triangle is a dual of the prime (original) triangle, but also that since
both the polar and the prime triangle are in the same space (triangles in a sphere) the operator dual here idempotent. That is, the square of the mapping $P$ of triangles to polar triangles is the identity $P^2=I$.  This is interesting for pure theoretical purposes.  
To better understand the duality we exploit the fact that a triangle, its polar image, and the relation between them is rotationally invariant. That is, when we want to understand properties for them we can choose a rotation (or coordinate system) that makes the problem easy. I include the following figure because :  what is easier that start with a 90-90-90 triangle with coordinates  at the three unit vectors (1,0,0), (0,1,0), and (0,0,1) and then transform this "unit" triangle into a more general triangle while observing the effect on the polar triangle? The first thing that we note is that a 90-90-90 triangle (equilateral) is equal to its polar triangle. This is easy to see because each vertex is the north pole of the equator passing through the other two vertices. This is shown in the green frame in the upper left figure.

We now want to deform or stretch this triangle to see how the dual (polar) triangle will change. If we keep the north pole at the angle $\gamma$ fixed and start with a wider triangle with central (angles at the center of the sphere) angles $-20^{\circ}$ to
$110^{\circ}$ in the base (plane $XY$), we see that the dual (the prime is red, the dual is blue) narrows to an angle $20-70$ (central angles have the same measures of arcs in unit spheres). The prime $\gamma$ angle in the upper part of the $Z$ axis is $\gamma = 110-(-20)= 130^{\circ}$. Still $\alpha=\beta=90^{\circ}$ since they are the angles between an equator and a parallel line with the constant azimuth (or constant latitude), however the
angle at $\gamma=110^{\circ}$ has the effect of squeezing the arc $c'$ .  Then $c'=70-20=50$, so
$\gamma + c' =\pi$. Likewise red arc at the bottom has $110-(-20)=130^{\circ}$,
and the blue angle at the top has $70-20=50^{\circ}$. The symbols in white background boxes mean no change. It does not matter if the shift was done by $20^{\circ}$ on each direction or any other amount. The idea is that the sum of the angles/arcs $\gamma+c'=\gamma'+c=\pi$. We could do the same experiment assuming that the  angle $\alpha$ is our north pole and the equator is in the plane with the $\beta$, $\gamma$ angles, and get 
$\alpha+a'=\alpha'+a=\pi$. The frame in the bottom consider two levels of deformation: in the horizontal an in the vertical $X-Z$ planes. The invariance under rotation implies that the equations:
\begin{eqnarray}
   \alpha + a' = \alpha'+a = \beta + b' = \beta'+b = \gamma + c' = \gamma' + c = \pi
\end{eqnarray}
are a constant. That is the sum of the angles with the polar arcs and viceversa (arcs with polar angles) is an invariant of the problem with constant $\pi$.
Again, properties such as fixed point objects, that is objects $X$ such
that if $P$ is the map between a triangle and its polar image remain fixed, or in equations $P(X)=X$, exists and these are all equilateral 90-90-90 triangles.
Applications
Linear Algebra:
The concept of dual basis is of great importance in linear algebra.
We now formulate the definition of a polar triangle in terms of linear algebra:
Definition: [polar triangle]
Given three unit vectors $A,B$, and $C$ in $\mathbb{R}^3$, linearly independent they form a basis for $\mathbb{R}^3$. We can build a "dual" basis as follows:
\begin{equation}
  A' = \frac{B \times C}{\| B \times C \|} \quad
  B' = \frac{C \times A}{\| C \times A \|} \quad
  C' = \frac{A \times B}{\| A \times B \|} \quad
\end{equation}
the new basis $\{A', B', C' \}$ constitutes a polar triangle.
Please observe the quotes around the word "dual" . They are dual
in the sense of what I describe here  . They have the direction of
the dual vectors in linear algebra. However the polar triangle (being in the unit sphere) has the vectors normalized. It is not hard to compute the the scaling between the polar (vectors) triangle and the dual base vectors is $s_k = \| X_i \times X_j \|/V$ where $V=X_1 \times X_2 \cdot X_3$ is
the volume of the parallelepiped formed by the three vectors $X_1,X_2$, 
and $X_3$ in the set $\{A,B,C\}$, and $i,j,k \in \{1,2,3\}$ and not repeated.
This scaling factors $s_k$, $k=1,2,3$ link the polar triangle with the dual basis of linear algebra. The fact that any north pole is orthogonal to vectors in the equator means that the matrix form by $[A, B, C]$ is
orthogonal to the matrix $[A', B', C']$ and so their product is diagonal. The diagonal is not the constant 1 due to the scaling factors to keep the polar triangle tight into a unit sphere.
Tensor Analysis:
In the field of tensors we have orthogonal coordinates which transform an space with no stretching or shear deformation, but we have general transformation which are no necessarily orthogonal. When the coordinates are orthogonal the concept of "covariant" or "contravariant" merge into one and the notation using sub-indices and super-indices is superfluous. This corresponds in spherical geometry to equilateral triangles where their polar duals are the same. In a general transformation which is not orthogonal we have two systems of coordinate axes. One is a covariant system and its (dual) is a contravariant. You will find that the axis of one system are those of a prime triangle (contravariant) while the axis of 
the other (the covariant ) are given by the polar triangle. Here the notation such as $\delta_{ijk}^{pql}$ with subindices and superindices is required.  Please beware of the scaling factors in the section of Linear Algebra above.
Functional Analysis:
In the field of functional analysis: A space and its dual (the dual is the space of functionals --functions from the space to the field $\mathbb{R}$ or $\mathbb{C}$) are such that coordinate axes in one space (say covariant for example) map into 
coordinate axes in the other space (contra-variant) in a way that correspond to the geometry of the angles (reducing the problem to $\mathbb{R}^3$ as a prototype) between the prime and the polar triangles respectively. Again, beware of scaling factors as explained above.
I have not seen this published but your question made me thing this.
Thanks.
A: Consider the ${fundamental} $ ${equations}$ of the spherical trigonometry, see figure, 
Let $A, B, C$ be three angles of a spherical triangle $ABC$; $a, b, c$, the sides opposite these vertices respectively, then it can be shown that we can obtain three equations describing relationship between them, 
$$\cos a= \cos b \cos c + \sin b \sin c \cos A$$
$$\cos b= \cos c \cos a + \sin c \sin a \cos B$$
$$\cos c= \cos a \cos b + \sin a \sin b \cos C$$
These equations are the fundamental equations since given three quantities, the other three can be computed easily by solving above equations. 
By definition when two triangles are so related that the vertices of the one are the poles of the sides of the other, i.e. the relation of the triangle to another triangle is reciprocal, that triangle is called polar triangle. 
Now for the above figure, let $a',b',c'$ be the sides and $A',B',C'$ be the opposite angles of the polar triangle corresponding to the original triangle $ABC$ such that:
$$a' = 180 - A, A' = 180-a$$ 
$$b' = 180 - B, B' = 180-b$$ 
$$c' = 180 - C, C' = 180-c$$ 
Since the fundamental equations are also satisfied by the polar triangle, using which new relations can be established from the fundamental equations about the original fundamental quantities:
$$\cos A = - \cos B \cos C + \sin B \sin C \cos a$$
$$\cos B = - \cos C \cos A + \sin C \sin A \cos b$$
$$\cos C = - \cos A \cos B + \sin A \sin B \cos c$$
Now to answer your question "why (and when) it is easier working with polar", you can see that in the above derivation, the use of polar triangle is not necessary. They can be directly derived from the fundamental equations by elimination of first $b$ and $c$, then $c$ and $a$, then $a$ and $b$. But the use of the polar triangle gives a more intuitive geometrically assisted, shorter and supposedly more elegant mode of deducing them.  Because of the fact that the sides and the angles of a triangle are respectively supplementary to the angles and sides of the polar triangle, many relations as such can be easily derived using the Polar triangle, hence a triangle and its polar are sometimes called "supplemental triangles."
You can refer to Spherical Trigonometry, Spherical Trigonometry For Colleges for more details.
