Question

prove : tan theta+ 2 tan 2 theta+4tan 4 theta + 8 cot 8 thetha = cot thetha​

Answer 1

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

Consider,

$\rm :\longmapsto\:\sf \: tan\theta + 2tan2\theta + 4tan4\theta + 8cot8\theta$

We know,

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

So,

$\boxed{ \tt{ \: cot2x = \frac{1 - {tan}^{2} x}{2tanx} \: \: }}$

So, using this, we get

$\sf \:=tan\theta + 2tan2\theta + 4tan4\theta + 8 \times \bigg[\dfrac{1 - {tan}^{2} 4\theta }{2tan4\theta } \bigg]$

$\sf \:=tan\theta + 2tan2\theta + 4tan4\theta + 4\times \bigg[\dfrac{1 - {tan}^{2} 4\theta }{tan4\theta } \bigg]$

$\sf \:=tan\theta + 2tan2\theta + 4tan4\theta + \bigg[\dfrac{4 - 4{tan}^{2} 4\theta }{tan4\theta } \bigg]$

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

$\sf \:=tan\theta + 2tan2\theta + \bigg[4tan4\theta + \dfrac{4 - 4{tan}^{2} 4\theta }{tan4\theta }\bigg]$

$\sf \:=tan\theta + 2tan2\theta + \bigg[\dfrac{4 {tan}^{2}4\theta + 4 - 4{tan}^{2} 4\theta }{tan4\theta }\bigg]$

$\sf \:=tan\theta + 2tan2\theta + \bigg[\dfrac{4}{tan4\theta }\bigg]$

$\sf \:=tan\theta + 2tan2\theta + \bigg[\dfrac{4(1 - {tan}^{2} 2\theta )}{2tan2\theta }\bigg]$

$\sf \:=tan\theta + 2tan2\theta + \bigg[\dfrac{2(1 - {tan}^{2} 2\theta )}{tan2\theta }\bigg]$

$\sf \:=tan\theta + 2tan2\theta + \bigg[\dfrac{2 - 2{tan}^{2} 2\theta}{tan2\theta }\bigg]$

$\sf \:=tan\theta + \bigg[2tan2\theta + \dfrac{2 - 2{tan}^{2} 2\theta}{tan2\theta }\bigg]$

$\sf \:=tan\theta + \bigg[\dfrac{ {2tan}^{2}2\theta + 2 - 2{tan}^{2} 2\theta}{tan2\theta }\bigg]$

$\sf \:=tan\theta + \bigg[\dfrac{ 2}{tan2\theta }\bigg]$

$\sf \:=tan\theta + \bigg[\dfrac{ 2(1 - {tan}^{2} \theta )}{2tan\theta }\bigg]$

$\sf \:=tan\theta + \bigg[\dfrac{ (1 - {tan}^{2} \theta )}{tan\theta }\bigg]$

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

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

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

$\sf \:= \: cot\theta$

Hence,

$\: \: \boxed{ \tt{ \: \sf \: tan\theta + 2tan2\theta + 4tan4\theta + 8cot8\theta = cot\theta}}$

Proved

Additional Information :-

$\boxed{ \tt{ \: sin2x = 2sinxcosx = \frac{2tanx}{1 + {tan}^{2}x } \: \: }}$

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

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

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

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

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

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

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