tan a = b/c

prove that:

c*cos2a + b sin2a = c

We know that :

sin 2a = 2tan a / (1+ tan^2 a )

==> b sin2a = b*[2tan a / (1+ tan^2 a]

cos2a = (1-tan^2 a)/(1+ tan^2 a)

==> c*cos2a = c*(1- tan^2a) /(1+ tan^2 a)

c*cos2a + b*sin2a = c(1-tan^2 a)/(1+ tan^2 a) + b[2tan a/ (1+ tan^2 a)

= [c(1-tan^2 a) + b *2tana)]/ (1+ tan^2 a)

But tan a = b/c

==> c(1-b^2/c^2 + 2b*b^2/c^2)/ (1+ b^2/c^2)

==> (c - b^2/c + 2b^3/c^2) / (1= b^2/c^2)

==> c ( 1+ b^2/c^2) / (1+ b^2/c^2)

==> c*1 = c

Give tan a = b/c. Then to prove that ccos2a =+bsin2a = c.

We know that tan a = b/c.

sina = tana /sqrt(1+tan^2) =( b/c)/sqrt{1+(b/c)^2} = b/sqrt(b^2+c^2).........(1)

Similarly,

cosa = c/sqrt(b^2+c^2)........(2)

sin2a = 2sina*cosa is an identity

sin2a = 2[b*/sqrt(b^2+c^2)]{c/sqrt(b^2+c^2)] using values at (1) and (2).

sin2a = 2bc/(b^2+c^2).........(3)

cos2a = cos^2 a - sin^2a is an identity.

cos2A = c^2 /(b^2+c^2)- b^2/(b^2+c^2)... (4)

ccos2a +bsin2a = c {c^2 -b^2)/(b^2+c^2) +b(2bc)/(b^2+c^2)

= c (c^2-b^2+2b^2)/(b^2+c^2)

=c(c^2+b^2)/(b^2+c^2)

=c.

cos2a =

We'll write cos 2a = [1 - (tan a)^2]/[1 + (tan a)^2] and

sin 2a = 2 tan a/[1 + (tan a)^2]

c*cos2a + b*sin2a = c* [1 - (tan a)^2]/[1 + (tan a)^2] + b*2 tan a/[1 + (tan a)^2]

We'll have the same denominator, so we could re-write:

c*cos2a + b*sin2a = c* [1 - (tan a)^2]b*2 tan a/[1 + (tan a)^2]

But tan a = b/c

c*cos2a + b*sin2a = c*[1 - (b/c)^2]b*2 *b/c/[1 + (b/c)^2]

c*cos2a + b*sin2a = c*(c^2 + b^2)/(c^2 + b^2)

We'll cancel the like terms from denominator and nnumerator:

**c*cos2a + b*sin2a = c**