Question

tan x + 2tan 2x +4 tan 4x + 8tan 8x = ?

Answer 1

$\underline{\bold{Simplification :}}$

We know that, $tan2x =\frac{2tanx}{1-tan^{2}x}$

So, $cot2x = \frac{1-tan^{2}x}{2tanx}$ ...(a)

We find, $tanx - cotx$

= $tanx - \frac{1}{tanx}$

= $\frac{tan^{2}x-1}{tanx}$

= $- 2*\frac{1-tan^{2}x}{2tanx}$

= $- 2cot2x$ ...(i) , by (a)

Similarly,

$tan2x - cot2x = - 2 cot4x$ ...(ii)

$tan4x - cot4x = - 2 cot8x$ ...(iii)

$tan8x - cot8x = - 2 cot16x$ ...(iv)

Now, $\small{\bold{tanx + 2tan2x + 4tan4x + 8tan8x}}$

= (tanx - cotx) + 2tan2x + 4tan4x + 8tan8x + cotx

= - 2cot2x + 2tan2x + 4tan4x + 8tan8x + cotx , by (i)

= 2 (tan2x - cot2x) + 4tan4x + 8tan8x + cotx

= 2 (- 2cot4x) + 4tan4x + 8tan8x + cotx , by (ii)

= 4 (tan4x - cot4x) + 8tan8x + cotx

= 4 (- 2cot8x) + 8tan8x + cotx , by (iii)

= 8 (tan8x - cot8x) + cotx

= 8 (- 2cot16x) + cotx , by (iv)

= $\boxed{\bold{- 16cot16x + cotx}}$ ,

which is the required simplified value.

Answer 2

Answer:

-16 cot16x +cotx

Step-by-step explanation:

We  know that,

tanx - cotx = sinx/cos - cosx/sinx

                 = (sin²x - cos²x)/sinx.cosx

                 = -(cos²x - sin²x)/sinx.cosx

                 = -2(cos²x - sin²x)/2sinx.cosx

                 = -2(cos2x)/sin2x

tanx- cotx = -2cot2x

This is an identity, because it  is  true for any value of  x .

(Note:-

Putting x = 2x, We  get,

tan2x - cot2x = -2cot4x

Putting x = 4x , We gate

tan4x - cot4x = -2cot8x

Similarly for  more.

We  have,

tanx + 2tan2x + 4tan4x + 8tan8x  

(tanx -cotx) + 2tan2x + 4tan4x + 8tan8x  + cotx

= -2cot2x + 2tan2x + 4tan4x + 8tan8x  + cotx

= 2(tan2x - cot2x) + 4tan4x + 8tan8x  + cotx

= 2.(-2cot4x) + 4tan4x + 8tan8x  + cotx

= -4cot4x + 4tan4x + 8tan8x  + cotx

= 4(tan4x - cot4x) + 8tan8x  + cotx

= 4(-2cot8x) + 8tan8x  + cotx

= -8cot8x +8tan8x + cotx

= 8(tan8x-cot8x) + cotx

= 8(-2cot16x) + cotx

= -16cot16x +cotx