Let's Start By Expanding The Expression:- #csc(u-v)#.
So, We have,
#csc(u - v)#
#= 1/sin(u - v)#
#= 1/(sin u cos v - cos u sin v)#.................(i)
[Expanded using the Identity #sin(A - B) = sinAcosB - cosAsinB#]
Now, We have to find #sin v# and #cos u# too.
We all know,
#color(white)(xxx)sin^2theta + cos^2theta = 1#
#rArr sin^2theta = 1 - cos^2 theta#
#rArr sin theta = +- sqrt(1 - cos^2 theta)#
And, Vice Versa.
So, #sin v = +-sqrt(1 - cos^2 v) = +-sqrt(1 - (-3/5)^2) = +-sqrt((25 - 9)/25) = +-4/5#
And, #cos u = =-sqrt(1 - sin^2 u) = +-sqrt(1 - (5/13)^2) = +-sqrt((169 - 25)/169) = +-12/13#
As, #sin v# and #cos u# have both positive and negative values, there will be FOUR values (maybe different or same) [as #4# combination #2 = 4#] for #csc(u - v)#.
So,
From (i),
First Value for #csc(u - v) = 1/((cancel5/13 xx-3/cancel5 )- (4/5 xx 12/13)) = 1/(-3/13 - 48/65) = 1/(-(15 + 48)/65) = -65/63#
Second Value for #csc(u - v) = 1/((cancel5/13 xx-3/cancel5 )- (4/5 xx -12/13)) = 1/(-3/13 + 48/65) = 1/((48 - 15)/65) = 65/33#
Third Value for #csc(u - v) = 1/((cancel5/13 xx-3/cancel5 )- (-4/5 xx 12/13)) = 1/(-3/13 - 48/65) = 1/(-(15 + 48)/65) = -65/63#
Fourth Value for #csc(u - v) = 1/((cancel5/13 xx-3/cancel5 )- (-4/5 xx -12/13)) = 1/(-3/13 + 48/65) = 1/((48 - 15)/65) = 65/33#
We can see First and Third and Second and Fourth values for #csc(u -v)# are same.
So, Basically there are two values for #csc (u - v)#.
In the End,
#csc(u - v) = -65/63, 65/33# is #sin u = 5/13# and #cos v = -3/5#.
Hope this helps.
