Question

a = btanx, find asin2x + bcos2x

Answer 1

Answer:

Step-by-step explanation:

a = btanx

tanx = a/b

tan2x = $(2tanx)/(1-tan^2x)$

         = $(2a/b)/(1-a^2/b^2)$

         = $2ab/b^2-a^2$

∴ sin2x =  $(2ab)/\sqrt{(2ab)^2 + (b^2 - a^2)^2}$

            =   $(2ab)/(a^2+b^2)$

and cos2x = $(b^2 - a^2)/\sqrt{(2ab)^2 + (b^2 - a^2)^2}$

                   = $(b^2 - a^2)/(b^2 + a^2)$

So, asin2x + bcos2x

   = $(2a^2b)/(a^2+b^2) + (b^3 - a^2b)/(b^2 + a^2)$

   = $[b(a^2 + b^2)]/(a^2 + b^2)$

   = b.

Hope it helps!!!!