Prove: tan4A = 4tanA - 4tan^3A / 1 - 6tan^2A + tan^4A?

Question

Prove: tan4A = 4tanA - 4tan^3A / 1 - 6tan^2A + tan^4A?

Answer 1

1st use  the  sin and cos addition formulae

sin 2A = 2 sin A cos A       cos 2A = ....

plus the fact that sin^2 + cos^2 =1

to prove that tan 2A = (1- tan^2A) / (1 + tan^2 A)

Now apply this to tan 4A = ...in terms of tan 2A = .... in terms of tanA

and that's your answer.

Answer 2

Take the LHS

We know tan 2x = 2tanx/( 1-tan²x)

Using this we can write,

tan 4A = tan2(2A)

          

           = 2 tan2A/( 1- tan² 2A)............(1)

  

tan2A = 2tanA / ( 1- tan² A)...............(2)

Putting (2) in (1)



               2 [2tanA / (1- tan² A)]                                                        

=-------------------------------------...                                    

              1- [ 2 tanA / (1- tan²A)]²

                 4 tanA /( 1- tan²A)                                                        

         =   ----------------------------------------...                                                

              [ (1- tan²A)² - 4tan²A] /( 1-tan²A)²                                        

[Cross multiply in denominator , now one of 1- tan²A gets cancelled]

                          4tan A                                                                  

         =     ----------------------------------------...

                [1 + tan^4 x - 2tan² A- 4tan²A]/(1-tan²A)

Denominator of the denominator is the numerator

         = 4tanA ( 1- tan²A)/[1 - 6tan²A + tan^4A]

        

         = (4 tanA - 4 tan³A) /[ 1- 6tan²A + tan^4A]

Sorry the equal sign is coming inside the fraction how many ever times i try