You can solve both by LH rule ... but lets try a different tune. First problem xa (sin-1x - sin-1a) / (x - a) =( - )/(sin - sin ) where sin-1x = and sin-1a = =( - )/2cos[( +)/2]sin[( -)/2] =(( - )/2)/sin[( -)/2] x (2/cos[( +)/2]) =1 x 2 =2
Second problem x/6 sin (x - /6) / 3 - 2cos x =x/6 sin (x - /6) / 2( 3/2 - cos x) =x/6 sin (x - /6) / 2(cos/6 - cos x) = x/6 2sin [(x - /6)/2] cos[(x - /6)/2]/ 4sin [(x + /6)/2]sin [(x - /6)/2] =x/6 (1/2)[cos[(x - /6)/2]/sin [(x + /6)/2]] =(1/2) x 1 / (1/2) = 1
Moderation Team
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5
a sin-1
x -sin-1
/6 sin(x-
3 - 2cosx (root sign is only for 3)
a (sin-1x - sin-1a) / (x - a)
(
and sin-1a = 
/6 sin (x - 
3/2 - cos x)
