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

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.