Question

sinAsin2A+sin3Asin6A+sin4Asin13A/sinAcos2A+sin3Acos6A+sin4Acos13A = tan9A

Answer 1

This one requires a bit of ingenuity, and a good knowledge of the sum/difference/product formulas.

sin A sin B = ½ cos(A−B) − ½ cos(A+B)
sin A cos B = ½ sin(A+B) + ½ sin(A−B)
so,
sinAsin2A = ½cosA - ½cos3A
sin3Asin6A = ½cos3A - ½cos9A
sin4Asin13A = ½cos4A - ½cos17A
add 'em up to get
½cosA - ½cos17A

Do likewise with the denominator to get

½sin17A - ½sinA

Now we have

(cosA - cos17A)/(sin17A - sinA)

We turn now to the other identities:

cos u − cos v = −2 sin(½(u+v)) sin(½(u−v))
sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))

That leaves us with

-2 sin9A sin-8A
----------------
2 sin8A cos9A

= tan 9A

Answer 2

oops.

sin4Asin13A = ½cos4A - ½cos17A
should read
sin4Asin13A = ½cos9A - ½cos17A