# Given sin x + cos x = 1, what is tan(x+x)?

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We have to find tan(x + x), given that sin x + cos x = 1.

We know that (sin x)^2 + (cos x)^2 = 1

sin x + cos x = 1

=> (sin x + cos x)^2 = 1

=> (sin x)^2 + (cos x)^2 + 2 sin x cos x = 1

=> 1 + 2 sin x cos x = 1

=> 2 sin x cos x = 0

=> sin 2x = 0

Now tan (x +x) = tan 2x = sin 2x / cos 2x

as sin 2x = 0

tan 2x = sin 2x/ cos 2x = tan (x +x) = 0

**Therefore we prove that tan (x +x) = 0**

We'll start by square raising the constraint from enunciation:

sinx + cosx = 1

(sinx + cosx)^2 = 1^2

(sinx)^2 + (cosx)^2 + 2sinx*cosx = 1 (1)

But, from the fundamental formula of trigonometry:

(sinx)^2 + (cosx)^2 = 1

We'll substitute (sinx)^2 + (cosx)^2 by 1:

The relation (1) will become:

1 + 2sinx*cosx = 1

We'll eliminate like terms:

2sinx*cosx = 0

But 2sinx*cosx = sin (2x)

We'll write the formula for tan(x+x) = tan 2x:

tan 2x = sin 2x/cos 2x

Since sin 2x = 0, we'll get:

tan 2x = 0/cos 2x

**tan (x+x) = tan 2x = 0**