# What is tan 2a when sin a +cos a = 1 ?

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It is given that sin a + cos a = 1

sin a + cos a = 1

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

(sin a)^2 + (cos a)^2 +2*sin a*cos a = 1

=> 1 + 2*sin a*cos a = 1

=> 2*sin a*cos a = 0

tan 2a = sin 2a / cos 2a

but sin 2a = 2*sin a*cos a = 0

=> tan 2a = 0

**The required value of tan 2a = 0**

First, we'll raise to square the given constraint:

sin a + cos a = 1

(sin a + cos a)^2 = 1

We'll expand the square:

(sin a)^2 + 2sin a*cos a + (cos a)^2 = 1

But, from Pythagorean identity, we'll get:

(sin a)^2 + (cos a)^2 = 1

1 + 2sin a*cos a = 1

We'll eliminate 1 both sides:

2sin a*cos a = 0

We recognize the double angle identity:

2sin a*cos a = sin (2a)

sin (2a)= 0

We'll write the tangent that has to be determined:

tan (2a) = sin (2a) /cos (2a)

tan (2a) = 0/cos (2a)

tan (2a) = 0

**The requested tangent is: tan (2a) = 0.**