Question 854271
Tan a=-4/3,  condition pi/2<a<pi   
Cos b = 1/2,  condition. 0<b<pi/2
Need exact value  cos(a+b).      
Sin(a-b).     
Tan(a-b).         

I believe  a would be in quadrant 2 where x is negative and y is positive. 
And b would bi in quadrant 1 where x and y are both positive.            
 How do I get tan b.  or do I need  sin a and cos a. If so how do I find them.
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tan(a) = -4/3, so y = 4 and x = -3
Then r = sqrt(16 + 9] = 5
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cos(b) = 1/2, so x = 1 and r = 2
Then y = sqrt(2^2-1) = sqrt(3)
And tan(b) = y/x = sqrt(3)/1 = sqrt(3)
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Need exact value  
cos(a+b) = cos(a)cos(b)-sin(a)sin(b) = (-3/5)(1/2)-(4/5)(sqrt(3)/2)
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Sin(a-b) = sin(a)cos(b)+cos(a)sin(b) = (4/5)(1/2)+(-3/5)(sqrt(3)/2)
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Tan(a-b) = [tan(a)+tan(b)]/[1-tan(a)tan(b)]
= [(-4/3)+sqrt(3)]/[1-(-4/3)sqrt(3)]
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Cheers,
Stan H.
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