mtan(thita+30) = ntan (thita+120 ) then m+n/m-n so how can we solve
Ritul , 8 Years ago
Grade 11
Mechanics> mtan(thita+30) = ntan (thita+120 ) then m...
2 AnswersAmrit Raj
Vikas TU
- m tan(Ѳ+30) = n tan(Ѳ+120)
- m/n = tan(Ѳ+120)/ tan(Ѳ+30)
- m/n = ((tanѲ+ tan120)/(1- tanѲ tan120))/(( tanѲ +tan30)/(1- tanѲ tan30))
- m/n =(( tanѲ -√3)/(1+√3tanѲ))/((√3tanѲ-1)/(√3- tanѲ))
- m/n=(( tanѲ -√3)* (√3- tanѲ))/( (1+√3 tanѲ)* (√3tanѲ-1))
- =((tanѲ -√3)2/(1√3 tanѲ)2)
- (tan2Ѳ +3-2√3 tanѲ)/(1-3tan2Ѳ)
- m/n= (tan2Ѳ +3-2√3 tanѲ)/(1-3tan2Ѳ)
Applying componendo and dividend
m+n/m-n = (-2 tan2Ѳ+4-2√3 tanѲ)/(2(1-3tan2Ѳ))
m+n/m-n =(- tan2Ѳ+2 -√3 tanѲ)/( 1-3tan2Ѳ)
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