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tan​theta = sina-cosa/sina+cosa, then sina +cosa is

tan​theta = sina-cosa/sina+cosa, then sina +cosa is

Grade:12th pass

2 Answers

Deepak
13 Points
4 years ago
Tanθ= (Sinα-Cosα)/(Sinα+Cosα)Now squaring both sidesTan2θ= (Sinα-Cosα)2/(Sinα+Cosα)2We know that Sin2α+Cos2α=1 soTan2θ= (1-2SinαCosα)/(1+2SinαCosα)Now add 1 both sides to use this identity Tan2θ+1= Sec2θTan2θ+1=(1-2SinαCosα)/(1+2SinαCosα) +1Take LCMSec2θ= (1-2SinαCosα+1+2SinαCosα)/(1+2SinαCosα)Now after subtractingSec2θ=2/(1+2SinαCosα)(1+2SinαCosα)=2/Sec2θNow we can write 1 as Sin2α+Cos2αSo it becomes (Sinα+Cosα)2=2/Sec2θWhere Sec2θ=1/Cos2θSo(Sinα+Cosα)2=2Cos2θTaking root both sides Sinα+Cosα=√2CosθAnswer.
Sagar
13 Points
one year ago
TanA=sinA-cosA/sinA+cosA
SinA/cosA=1-cosA-cosA/sinA+cosA
SinA/cosA=1-2cosA/sinA+cosA
1-cosA/cosA=cos2A/sinA+cosA
1-cosA/cosA=2cos^2-1/sinA+cosA
2-cosA/cosA=2cos^2/sinA+cosA
2-1=2cos^2/sinA+cosA
 
1=2cos^2/sinA+cosA
SinA+cosA=2cos^2
 
 
 
 
 
 
2-coA/

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