# Tan a – tanb = x, cotb -cota = Y,THEN cot (a-b)=

Y RAJYALAKSHMI
45 Points
9 years ago
we have tan A – tanB = sin (A – B) / cosA cosB  & cot B – cot A = sin (A – B)/ sinA sin B
1/x + 1/y = cosA cosB /sin(A – B) + sinA sinB / sin (A – B)
=(cosA cosB + sinA sinB) /sin (A – B)
= cos (A – B) / sin (A – B) = cot (A – B)

Ans:  cot (A – B) = 1/x + 1/y

Shriyans Naik
20 Points
6 years ago
tan A - tan B = x 1 1——— - ——— = xcot A cot Bcot B - cot A—————— = x(cot A)(cot B) y—————— = x(cot A)(cot B)(cot A)(cot B) = y — xcot (A-B) = (cot A)(cot B) + 1 ———————— cot B - cot A y = — + 1 ÷ x x x + y 1 = ——— × —— x x = x + y ——— x^2
Shriyans Naik
20 Points
6 years ago
tan A - tan B = x1/ cot A - 1/ cot B = x(cot B - cot A) / (cot A)(cot B) = xy / (cot A )( cot B ) = x(cot A)(cot B) = y / xcot (A - B) =[ (cot A)(cot B) + 1] /cot B -cot Acot (A - B) =[( y / x) + 1] / y cot (A - B)= [ (y+x) / x ] / ycot (A-B) = (y + x) / (xy)cot (A - B)=( y / xy) + ( x /xy)cot (A- B) = (1/y) +(1/x)
3 years ago
Dear student,

tanA – tanB = x
cotB – cotA = y
Hence, 1/tanB – 1/tanA = y
or, (tanA – tanB)/tanAtanB = y
or, tanAtanB = x/y

Now, tan(A – B) = (tanA – tanB)/(1 + tanAtanB)
= x/(1 + x/y)
= xy/(x + y)
Hence, cot(A – B) = 1/tan(A – B)
= (x + y)/xy = 1/x + 1/y

Thanks and regards,
Kushagra