# Determine tan2a if sina + cosa = 1Determine tan2a if sina + cosa = 1

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sina + cosa = 1

==> sina = 1-cosa

We know that:

tan2a = 2tana / [1-(tana)^2]

But tana = sina/cosa

==> tan2a = 2(sina/cosa) / [1- (sina/cosa)^2]

= 2(1-cosa)/cosa / [1-(1-cosa/cosa)^2

= 2(1-cosa)/cosa / [cos^2 - (1-cosa)^2]/cos^2 a)

= 2(1-cosa)*cosa/ (cos^2 a - (1-2cosa +cos^2)

= 2cosa(1-cosa)/(2cosa -1)

We know that for any angle a (sin a)^2+(cos a)^2=1.

Here sin a+cos a=1. Take the square of both the sides, we get

(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

So either sin a is 0 or cos a is 0

If sin a is 0, a=0, 2a=0, cos 0=1 Therefore tan 2a=1/0 is infinity.

If cos a =0, a= pi/2, 2a=pi, cos pi=-1, sin pi= 0 so tan 2a=0

Therefore tan 2a is either 0 or infinity.

Given sina +cosa = 1.

We know that s^2a+cos^2a =1. So cosa = sqrt(1-sin^2).

Threfore sina +sqrt(1-sin^2a) =1

1-sina = -sqrt(1-sin^2a). square both sides:

1-2sina +sin^2a = 1-sin^2

-2sina+2sin^2a = 0

-sina(1-sina) =0.

sina = 0 or sina = pi/2

When sina =0, cosa = 1 in first quadrant . So tana = sina/cosa = 0. In 2nd (or 3rd quadrant) also sina/cosa = 0.

When sina =1,

cosa = 1 in 1st quadrant and cosa = 0+ tana = 1/o+ = positive inf.

In 2nd quadrant, cosa = 0- So tana =1/0- = -ve inf.

We'll factorize the expression sina + cosa = 1 by cos a:

cos a*(sina/cosa + 1) = 1

We'll divide both sides by cos a:

sin a/cos a + 1= 1/cos a

tg a + 1= 1/cos a

tg a= 1/cos a - 1

But tg 2a could be written:

tg 2a=tg (a+a)=(tg a+ tga)/(1-(tga)^2)

tg 2a= 2tga/(1-(tga)^2)

tg 2a=2(1/cosa - 1)/[1-(1/cosa)+1][1+ (1/cosa)-1]

tg 2a= 2[(1-cos a)/cosa ]/[(2 - 1/cosa)(1/cos a)]

**tg 2a= 2[(1-cos a)]/[(2 - 1/cosa)]**