Question

(1+sinA/cosA) + (cosB/1-sinB) = [2sinA-2sinB] / [ sin(A-B)+ cosA -cosB]

Answer 1

sum_(n = 1)^n cos(a + (n - 1)*b) = cos[a + (b/2)*(n - 1)]*sin(b*n/2)/sin(b/2) 

cos(a) + cos(a + b) + cos(a + 2b) + ... + cos(a + (n - 1)*b) = cos[a + (b/2)*(n - 1)]*sin(b*n/2)/sin(b/2) 

$\bf THEN \: WE \: CAN \: TYPE$


cos(a) + cos(a + b) + cos(a + 2b) + ... + cos(a + (n - 1)*b) + cos(a + b*n) = cos[a + (b*n/2)]* 
sin[b*(n + 1)/2]/sin(b/2) 

Remember that sum_(n = 1)^n cos(a + (n - 1)*b) = cos[a + (b/2)*(n - 1)]*sin(b*n/2)/sin(b/2) 

{cos[a + (b/2)*(n - 1)]*sin(b*n/2)/sin(b/2)} + cos(a + b*n) = cos[a + (b*n/2)]*sin[b*(n + 1)/2]/sin(b/2) 

{[cos(a)*[cos(b*n/2)*cos(b/2) + sin(b*n/2)*sin(b/2)] - sin(a)*[sin(b*n/2)*cos(b/2) - cos(b*n/2)*sin(b/2)]]* 
sin(b*n/2)/sin(b/2)} + cos(a)*cos(b*n) - sin(a)*sin(b*n) = 
[cos(a)*cos(b*n/2) - sin(a)*sin(b*n/2)]*[sin(b*n/2)*cos(b/2) + cos(b*n/2)*sin(b/2)]/sin(b/2) 

[{cos(a)*cos(b*n/2)*cos(b/2)*sin(b*n/2) + sin^2(b*n/2)*sin(b/2)*cos(a) - sin(a)*sin^2(b*n/2)*cos(b/2) + sin(a)*cos(b*n/2)*sin(b/2)*sin(b*n/2)}/s... + cos(a)*cos(b*n) - sin(a)*sin(b*n) = 
[cos(a)*cos(b*n/2) - sin(a)*sin(b*n/2)]*[sin(b*n/2)*cos(b/2) + cos(b*n/2)*sin(b/2)]/sin(b/2) 

[{cos(a)*cos(b*n/2)*cos(b/2)*sin(b*n/2) + sin^2(b*n/2)*sin(b/2)*cos(a) - sin(a)*sin^2(b*n/2)*cos(b/2) + sin(a)*cos(b*n/2)*sin(b/2)*sin(b*n/2)}/s... + cos(a)*cos(b*n) - sin(a)*sin(b*n) = 
[cos(a)*cos(b*n/2)*sin(b*n/2)*cos(b/2) + cos(a)*cos(b*n/2)*cos(b*n/2)*sin(b/2) - sin(a)*sin(b*n/2)* 
sin(b*n/2)*cos(b/2) - sin(a)*sin(b*n/2)*cos(b*n/2)*sin(b/2)]/s... 

[{sin^2(b*n/2)*sin(b/2)*cos(a) + sin(a)*cos(b*n/2)*sin(b/2)*sin(b*n/2)}/s... + cos(a)*cos(b*n) - sin(a)*sin(b*n) = [cos(a)*cos^2(b*n/2)*sin(b/2) - sin(a)*sin(b*n/2)*cos(b*n/2)*sin(b/2)]/s... 

[{sin^2(b*n/2)*sin(b/2)*cos(a) + sin(a)*cos(b*n/2)*sin(b/2)*sin(b*n/2)}/s... + cos(a)*cos(b*n) - sin(a)*sin(b*n) = {cos(a)*[1 - sin^2(b*n/2)]*sin(b/2) - sin(a)*sin(b*n/2)*cos(b*n/2)*sin(b/2)}/s... 

[{sin^2(b*n/2)*sin(b/2)*cos(a) + sin(a)*cos(b*n/2)*sin(b/2)*sin(b*n/2)}/s... + cos(a)*cos(b*n) - sin(a)*sin(b*n) = [cos(a)*sin(b/2) - sin^2(b*n/2)*sin(b/2)*cos(a) - sin(a)*sin(b*n/2)*cos(b*n/2)*sin(b/2)]/s... 

sin^2(b*n/2)*cos(a) + sin(a)*cos(b*n/2)*sin(b*n/2) + cos(a)*cos(b*n) - sin(a)*sin(b*n) = cos(a) - 
sin^2(b*n/2)*cos(a) - sin(a)*sin(b*n/2)*cos(b*n/2) 

2*sin^2(b*n/2)*cos(a) + sin(a)*cos(b*n/2)*sin(b*n/2) + cos(a)*cos(b*n) - sin(a)*sin(b*n) = cos(a) - sin(a)*sin(b*n/2)*cos(b*n/2) 

2*sin^2(b*n/2)*cos(a) + 2*sin(a)*cos(b*n/2)*sin(b*n/2) + cos(a)*cos(b*n) - sin(a)*sin(b*n) = cos(a) 

2*sin^2(b*n/2)*cos(a) + sin(a)*sin(b*n) + cos(a)*cos(b*n) - sin(a)*sin(b*n) = cos(a) 

2*sin^2(b*n/2)*cos(a) + cos(a)*cos(b*n) = cos(a) 

2*(1 - cos^2(b*n/2))*cos(a) + cos(a)*cos(b*n) = cos(a) 

2*cos(a) - 2*cos(a)*cos^2(b*n/2) + cos(a)*cos(b*n) = cos(a) 

2 - 2*cos^2(b*n/2) + cos(b*n) = 1 

2 - 2*cos^2(b*n/2) + cos^2(b*n/2) - sin^2(b*n/2) = 1 

2 - cos^2(b*n/2) - sin^2(b*n/2) = 1 

1 = 1 

Answer 2

d) If the circumradius of triangle ABC be R, then = a = b = c R 2sinA 2sinB 2sinC where a, b, ... then R = AC 2sinB = AB = AC Þ AC 2sinB = AC We sin know B= 1 2 that Þ ÐA sinB + ...