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
Last Activity: 7 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
Last Activity: 7 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)
Kushagra Madhukar
Last Activity: 4 Years ago
Dear student,
Please fin the solution to your problem.
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
Provide a better Answer & Earn Cool Goodies
Enter text here...
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free