Question

Pls help me out of this question $\large\purple{Given}$ cotA -tanA = 2cot2A then $\large\purple{To find}$ tanA - cotA +2tan2A + 4tan4A + 8cot8A No spamming ​

Answer 1

A N S W E R :

• tan A + 2tan2A + 4tan4A + 8cot8A = ${\sf{\dfrac{1}{tan A}}}$= cotA

Given :

• cotA -tanA = 2cot2A

To find :

• tanA - cotA +2tan2A + 4tan4A + ${\sf{\dfrac{8}{tan8A}}}$ ?

Solution :

${\sf{tan\: A\: + \:2tan2A \:+ \:4tan4A \:+ \; \dfrac{8\bigg(1-tan24A\bigg)}{2tan4A}}}$

${\sf{tan \:A \:+\: 2tan2A \:+}}$$\bigg[$$\bigg\{$${\sf{4tan4A\bigg(tan4A\bigg)}}$$\bigg\}$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$${\sf{\dfrac{+\: 4 \; \bigg(1-tan24A\bigg)}{tan4A}}}$

${\sf{\dfrac{tan \:A \:+ \:2tan2A\: +\; \bigg[4tan24A \;+ \;4\: -\; tan24A\bigg]}{tan4A}}}$

${\sf{\dfrac{tan \:A \:+ \;2tan2A \:+ \;4}{tan4A}}}$

Hence,

• ${\bf{\dfrac{tan \:A\: +\: 2tan2A\: + \:4tan4A \:+ \:8cot8A\: =\: 1}{tanA\: = \:cotA}}}$

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

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