Question

tan=sina-cosa/sina+cosa​

Answer 1

Answer:

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.

Answer 2

Tanx = (Sinα-Cosα)/(Sinα+Cosα)

Now squaring both sides

Tan^2x= (Sinα-Cosa)^2/(Sinα+Cosα)^2

We know that

Sin^2α+Cos^2α=1

so

Tan^2x= (1-2SinαCosα)/(1+2SinαCosα)

Now add 1 both sides

Tan^2x+1= Sec^2xTan^2x+1=(1-2SinαCosα)/(1+2SinαCosα) +1

Take LCM

Sec^2x= (1-2SinαCosα+1+2SinαCosα)/(1+2SinαCosα)

Now after subtracting

Sec^2x=2/(1+2SinαCosα)(1+2SinαCosα)=2/Sec2x

Now we can write 1 as Sin^2α+Cos^2α

So it becomes

(Sinα+Cosα)^2=2/Sec^2x

Where

Sec^2x=1/Cos^2c

So

(Sinα+Cosα)^2=2Cos^2x

Taking root both sides

Sinα+Cosα=√2Cosx Answer

Hope it's helpful for you