Question

tanx=sina-cosa\sina+cosa

Answer 1

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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

Answer 2

Answer:

answer for above question is the ❓

Root2