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Tan x +tan y=42 and cot x +cot y=49 then find the value of tan (x+y)

Abhay kumar , 12 Years ago
Grade 12
anser 2 Answers
Abhishek Doyal
tan(x+y)=[tan(x)+tan(y)]/1-tan(x)tan(y) eq 1
cot x +cot y = 1/tan x+1/tan y =(tan x +tan y)/tan (x) tan (y)
49=42/tan x.tan y
tan x.tan y=42/49=6/7 eq 2
by eq 1 and eq 2
tan(x+y)=42/(1-6/7)=7*42=294
Abhishek Doyal
IIT Bombay
Askiitian tutor

Last Activity: 12 Years ago
Varun Acharya
given: tanx + tany = 42 -------------- (1) cotx + coty = 49 ==> 1/tanx + 1/tany = 49 ==> (tanx + tany)/tan x tany = 49 hence tanx tany = (tanx + tany)/49, therefore tanx tany = 42/49 = 6/7--------------- (2) now tan(x+y) = (tanx + tany)/1-tanx.tany........ substituing from (1) and (2), we get tan(x+y) as 294....... is the answer correct??????
Last Activity: 12 Years ago
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