If tan a = b/c prove that c*cos2a + b*sin2a = 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
=