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
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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
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