Question

prove it
....

tan theta +2 tan 2 theta + 4 tan 4 theta + 8 cot 8 theta = cot theta​

Answer 1

$\underline{\bold{Given \:Question - }}$

Prove that

$\red{\rm :\longmapsto\:tan\theta + 2tan\theta + 4tan4\theta + 8cot8\theta = cot\theta }$

$\large\underline{\sf{Solution-}}$

Consider,

$\red{\rm :\longmapsto\:tan\theta + 2tan\theta + 4tan4\theta + 8cot8\theta }$

can be rewritten as

$\rm \: = \: \: tan\theta + 2tan2\theta + 4tan4\theta + \dfrac{8}{tan8\theta }$

We know,

$\boxed{ \rm{ tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}$

So, above can be rewritten as

$\rm \: = \: \: tan\theta + 2tan2\theta + 4tan4\theta + \dfrac{8}{\dfrac{2tan4\theta }{1 - {tan}^{2} 4\theta } }$

$\rm \: = \: \: tan\theta + 2tan2\theta + 4tan4\theta + \dfrac{4(1 - {tan}^{2} 4\theta )}{tan4\theta }$

$\rm \: = \: \: tan\theta + 2tan2\theta + \dfrac{4 {tan}^{2} \theta + 4(1 - {tan}^{2} 4\theta )}{tan4\theta }$

$\rm \: = \: \: tan\theta + 2tan2\theta + \dfrac{4 {tan}^{2} \theta + 4 - 4{tan}^{2} 4\theta }{tan4\theta }$

$\rm \: = \: \: tan\theta + 2tan2\theta + \dfrac{4 }{tan4\theta }$

Again, using

$\boxed{ \rm{ tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}$

$\rm \: = \: \: tan\theta + 2tan2\theta + \dfrac{4 }{\dfrac{2tan2\theta }{1 - {tan}^{2}2\theta } }$

$\rm \: = \: \: tan\theta + 2tan2\theta + \dfrac{2(1 - {tan}^{2} 2\theta )}{tan2\theta }$

$\rm \: = \: \: tan\theta + \dfrac{2 {tan}^{2}2\theta + 2 - 2{tan}^{2} 2\theta }{tan2\theta }$

$\rm \: = \: \: tan\theta + \dfrac{2}{tan2\theta }$

Again using,

$\boxed{ \rm{ tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}$

$\rm \: = \: \: tan\theta + \dfrac{2}{\dfrac{2tan\theta }{1 - {tan}^{2} \theta }}$

$\rm \: = \: \: tan\theta +\dfrac{1 - {tan}^{2} \theta }{tan\theta }$

$\rm \: = \: \: \dfrac{ {tan}^{2}\theta + 1 - {tan}^{2} \theta }{tan\theta }$

$\rm \: = \: \: \dfrac{1}{tan\theta }$

$\rm \: = \: \: cot\theta$

Hence, Proved

Additional Information :-

$\boxed{ \rm{ sin2x = 2sinx \: cosx}}$

$\boxed{ \rm{ sin2x = \frac{2tanx}{1 + {tan}^{2}x } }}$

$\boxed{ \rm{ cos2x = {cos}^{2}x - {sin}^{2} x}}$

$\boxed{ \rm{ cos2x = 1 - 2{sin}^{2} x}}$

$\boxed{ \rm{ cos2x = 2{cos}^{2} x - 1}}$

$\boxed{ \rm{ cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}}}$

$\boxed{ \rm{ sin3x = 3sinx - 4 {sin}^{3}x}}$

$\boxed{ \rm{ cos3x = 4 {cos}^{3}x - 3cosx}}$

Answer 2

Answer:

To prove:-

$\star \: \tt\tan \theta + 2\tan 2\theta + 4\tan 4\theta + 8\tan 8\theta = \cot\theta$

Step-by-step explanation:

We Know,

$: \longrightarrow \bf \cot \theta - \tan \theta = \frac{1}{\tan \theta} - \tan \theta \\ \\ \: \: \: \: : \longrightarrow \frac{1 - {\tan}^{2} \theta }{\tan \theta} = 2 \bigg\lgroup\frac{1 - {\tan}^{2} \theta }{\tan \theta}\bigg\rgroup$

$: \longrightarrow\cot\theta - \tan \theta = \frac{2}{\tan 2\theta} \\ \\ \longrightarrow\cot\theta - \tan \theta = 2\cot 2\theta \: ....(i)$

We have to prove that,

$\star \: \tt\tan \theta + 2\tan 2\theta + 4\tan 4\theta + 8\tan 8\theta = \cot\theta$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀or

$\star \: \tt\cot \theta - \tan \theta - 2\tan \theta - 4\tan 4\theta - 8\cot8\theta = 0$

Now, Taking left hand side (LHS),

$: \implies\cot \theta - \tan \theta - 2\tan \theta - 4\tan 4\theta - 8\cot8\theta \\ \\ : \implies(\cot \theta - \tan \theta ) - 2\tan \theta - 4\tan 4\theta - 8\cot8\theta \\ \\ : \implies2\cot 2\theta - 2\tan 2\theta - 4\tan 4\theta - 8\cot8\theta$

Using eqn (i)....

$: \implies \sf2(\cot 2\theta - \tan 2\theta) - 4\tan 4\theta - 8\cot8\theta \\ \\ : \implies \sf2(2\cot 4\theta) - 4\tan 4\theta - 8\cot8\theta \\ \\ : \implies \sf4\cot 4\theta - 4\tan 4\theta - 8\cot8\theta \\ \\ : \implies \sf4(\cot 4\theta - \tan 4\theta) - 8\cot8\theta \\ \\ : \implies \sf4(2\cot 8\theta ) - 8\cot8\theta \\ \\ : \implies \sf8\cot 8\theta - 8\cot8\theta \\ \\ : \implies0$

Hence , LHS = RHS