Question

If sinA=cosA then find tan2A

Answer 1

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)]

Answer 2

$sin\ A=cos\ A \\ \\ \frac{sin\ A}{cos\ A}=1\\ \\tan\ A =1\\ \\A=tan^{-1}1=45^0\\ \\2A=45^0*2=90^0\\ \\tan\ 2A=tan\ 90^0=\infty$