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2014-09-22T14:12:26+05:30

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TanA+ \frac{4}{TanA} =5
TanA = 5 -  \frac{4}{TanA}
TanA =  \frac{5TanA-4}{TanA}
 Tan^{2} A=5TanA-4
 tan^{2}A-5TanA+4=0
 Tan^{2} A-4TanA-TanA+4=0
TanA(TanA-4)-1(TanA-4)=0
(TanA-1)(TanA-4)=0
therefore TanA = 1
 And TanA = 4
When ,TanA = 1 then SinA = 1/√2   & CosA = 1/√2
And when TanA = 4 ,then SinA = 4/√17  & CosA = 1/√17
3 5 3
see whther mines is correct or not
IN WHICH DO YOU READ ?,IT'S COMPELETLY WRONG.
who said its wrong
sorry bro
It's ok